Limit Fungsi Aljabar
Limit Fungsi Aljabar
Matematika Kelas XI — Materi Lengkap, Contoh Soal & Latihan
1. Pengertian Limit Fungsi di Satu Titik
Limit fungsi \(f(x)\) saat \(x\) mendekati \(c\) adalah nilai yang didekati oleh \(f(x)\) ketika \(x\) semakin dekat ke \(c\), ditulis:
Artinya: semakin dekat \(x\) ke \(c\) (dari kiri maupun kanan), nilai \(f(x)\) semakin dekat ke \(L\).
Limit kiri: \(\displaystyle\lim_{x \to c^-} f(x)\) Limit kanan: \(\displaystyle\lim_{x \to c^+} f(x)\)
Limit ada jika dan hanya jika limit kiri = limit kanan.
Contoh Soal
Mudah1. Tentukan \(\displaystyle\lim_{x \to 2} (3x + 1)\)
Substitusi langsung: \(f(2) = 3(2)+1 = 7\)
Jadi \(\displaystyle\lim_{x \to 2}(3x+1) = 7\)
2. Tentukan \(\displaystyle\lim_{x \to 1} (x^2 + 2x)\)
Substitusi: \(1^2 + 2(1) = 3\)
Jadi limitnya = 3
3. Tentukan \(\displaystyle\lim_{x \to 0} (5x – 4)\)
Substitusi: \(5(0)-4 = -4\)
4. Tentukan \(\displaystyle\lim_{x \to 3} 7\)
Limit fungsi konstan selalu sama dengan konstantanya. Jadi = 7
5. Tentukan \(\displaystyle\lim_{x \to -1} (2x^2)\)
Substitusi: \(2(-1)^2 = 2\)
6. Tentukan \(\displaystyle\lim_{x \to 2} \frac{x^2-4}{x-2}\)
Substitusi langsung menghasilkan \(\frac{0}{0}\) (tak tentu).
Faktorkan: \(\frac{(x-2)(x+2)}{x-2} = x+2\)
Substitusi: \(2+2 = 4\)
7. Tentukan \(\displaystyle\lim_{x \to 3} \frac{x^2-9}{x-3}\)
Faktorkan: \(\frac{(x-3)(x+3)}{x-3} = x+3\)
Substitusi: \(3+3=6\)
8. Tentukan \(\displaystyle\lim_{x \to 1} \frac{x^2-1}{x^2-x}\)
\(\frac{(x-1)(x+1)}{x(x-1)} = \frac{x+1}{x}\)
Substitusi: \(\frac{2}{1}=2\)
9. Tentukan \(\displaystyle\lim_{x \to 4} \frac{x^2-16}{x^2-3x-4}\)
\(\frac{(x-4)(x+4)}{(x-4)(x+1)} = \frac{x+4}{x+1}\)
Substitusi: \(\frac{8}{5}\)
10. Tentukan \(\displaystyle\lim_{x \to -2} \frac{x^2+5x+6}{x+2}\)
\(\frac{(x+2)(x+3)}{x+2} = x+3\)
Substitusi: \(-2+3=1\)
11. Tentukan \(\displaystyle\lim_{x \to 1} \frac{x^3-1}{x^2-1}\)
\(\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{x^2+x+1}{x+1}\)
Substitusi: \(\frac{3}{2}\)
12. Tentukan \(\displaystyle\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}\)
Kalikan sekawan: \(\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}\)
Substitusi: \(\frac{1}{4} \)
13. Tentukan \(\displaystyle\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}\)
\(\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3} = \sqrt{x}+3\)
Substitusi: \(3+3=6\)
14. Tentukan \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x+1}-1}{x}\)
Kalikan sekawan: \(\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \frac{x}{x(\sqrt{x+1}+1)} = \frac{1}{\sqrt{x+1}+1}\)
Substitusi: \(\frac{1}{2}\)
15. Tentukan \(\displaystyle\lim_{x \to 2} \frac{x^3-8}{x^2-4}\)
\(\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2}\)
Substitusi: \(\frac{12}{4}=3\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x \to 5}(2x-3)\)
2. \(\displaystyle\lim_{x \to -2}(x^2+1)\)
3. \(\displaystyle\lim_{x \to 0}(4x+7)\)
4. \(\displaystyle\lim_{x \to 1}(x^3)\)
5. \(\displaystyle\lim_{x \to 2}(3x^2-x)\)
6. \(\displaystyle\lim_{x \to 5}\frac{x^2-25}{x-5}\)
7. \(\displaystyle\lim_{x \to -3}\frac{x^2-9}{x+3}\)
8. \(\displaystyle\lim_{x \to 2}\frac{x^2-3x+2}{x-2}\)
9. \(\displaystyle\lim_{x \to 1}\frac{x^3-x}{x-1}\)
10. \(\displaystyle\lim_{x \to 0}\frac{x^2+3x}{x}\)
11. \(\displaystyle\lim_{x \to 1}\frac{\sqrt{x}-1}{x-1}\)
12. \(\displaystyle\lim_{x \to 0}\frac{\sqrt{4+x}-2}{x}\)
13. \(\displaystyle\lim_{x \to 8}\frac{x-8}{\sqrt[3]{x}-2}\)
14. \(\displaystyle\lim_{x \to 3}\frac{x^3-27}{x^2-9}\)
15. \(\displaystyle\lim_{x \to 1}\frac{x^4-1}{x^3-1}\)
2. Sifat-sifat Limit Fungsi di Satu Titik
Jika \(\displaystyle\lim_{x\to c}f(x)=L\) dan \(\displaystyle\lim_{x\to c}g(x)=M\), maka:
- \(\displaystyle\lim_{x\to c}[f(x)\pm g(x)] = L \pm M\)
- \(\displaystyle\lim_{x\to c}[k \cdot f(x)] = k \cdot L\)
- \(\displaystyle\lim_{x\to c}[f(x)\cdot g(x)] = L \cdot M\)
- \(\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)} = \frac{L}{M},\; M\neq 0\)
- \(\displaystyle\lim_{x\to c}[f(x)]^n = L^n\)
- \(\displaystyle\lim_{x\to c}\sqrt[n]{f(x)} = \sqrt[n]{L},\; L\geq 0\)
Contoh Soal
Mudah1. Jika \(\lim_{x\to 2}f(x)=3\) dan \(\lim_{x\to 2}g(x)=5\), tentukan \(\lim_{x\to 2}[f(x)+g(x)]\)
Sifat penjumlahan: \(3+5=8\)
2. Tentukan \(\lim_{x\to 2}[3f(x)]\) jika \(\lim_{x\to 2}f(x)=4\)
Sifat perkalian konstanta: \(3\cdot4=12\)
3. \(\lim_{x\to 1}[f(x)\cdot g(x)]\) jika \(\lim f=2,\;\lim g=6\)
\(2\times6=12\)
4. \(\lim_{x\to 3}\frac{f(x)}{g(x)}\) jika \(\lim f=10,\;\lim g=2\)
\(\frac{10}{2}=5\)
5. \(\lim_{x\to 4}[f(x)]^2\) jika \(\lim f=3\)
\(3^2=9\)
6. Jika \(\lim_{x\to 1}f(x)=4,\;\lim_{x\to 1}g(x)=-1\), tentukan \(\lim_{x\to 1}[2f(x)-3g(x)]\)
\(2(4)-3(-1)=8+3=11\)
7. \(\lim_{x\to 2}\frac{f(x)+g(x)}{f(x)-g(x)}\) jika \(\lim f=5,\;\lim g=3\)
\(\frac{5+3}{5-3}=\frac{8}{2}=4\)
8. \(\lim_{x\to 0}\sqrt{f(x)}\) jika \(\lim f=16\)
\(\sqrt{16}=4\)
9. \(\lim_{x\to 1}[f(x)]^3\) jika \(\lim f=-2\)
\((-2)^3=-8\)
10. \(\lim_{x\to 2}[f(x)\cdot g(x)+g(x)^2]\) jika \(\lim f=3,\;\lim g=2\)
\(3\cdot2+2^2=6+4=10\)
11. Jika \(\lim_{x\to 1}f(x)=2\), tentukan \(\lim_{x\to 1}\frac{f(x)^3+f(x)}{f(x)^2-1}\)
\(\frac{8+2}{4-1}=\frac{10}{3}\)
12. \(\lim_{x\to 0}\frac{\sqrt{f(x)+9}-3}{f(x)}\) jika \(\lim f=0\). Gunakan substitusi \(u=f(x)\).
Kalikan sekawan: \(\frac{u}{u(\sqrt{u+9}+3)}=\frac{1}{\sqrt{u+9}+3}\)
Saat \(u\to0\): \(\frac{1}{3+3}=\frac{1}{6}\)
13. \(\lim_{x\to 2}\left[\frac{f(x)}{g(x)}\right]^2\) jika \(\lim f=6,\;\lim g=3\)
\(\left(\frac{6}{3}\right)^2=4\)
14. \(\lim_{x\to 1}\frac{f(x)^2-g(x)^2}{f(x)-g(x)}\) jika \(\lim f=5,\;\lim g=3\)
\(\frac{(f-g)(f+g)}{f-g}=f+g=5+3=8\)
15. \(\lim_{x\to 4}\frac{\sqrt{f(x)}-\sqrt{g(x)}}{f(x)-g(x)}\) jika \(\lim f=9,\;\lim g=4\)
\(\frac{\sqrt f-\sqrt g}{(\sqrt f-\sqrt g)(\sqrt f+\sqrt g)}=\frac{1}{\sqrt f+\sqrt g}=\frac{1}{3+2}=\frac{1}{5}\)
Latihan Soal
Mudah1. \(\lim_{x\to3}[f(x)-g(x)]\) jika \(\lim f=7,\;\lim g=4\)
2. \(\lim_{x\to1}[5f(x)]\) jika \(\lim f=2\)
3. \(\lim_{x\to0}[f(x)\cdot g(x)]\) jika \(\lim f=3,\;\lim g=4\)
4. \(\lim_{x\to2}\frac{f(x)}{g(x)}\) jika \(\lim f=8,\;\lim g=4\)
5. \(\lim_{x\to1}[f(x)]^3\) jika \(\lim f=2\)
6. \(\lim_{x\to1}[3f(x)+2g(x)]\) jika \(\lim f=4,\;\lim g=-3\)
7. \(\lim_{x\to0}\sqrt{f(x)+g(x)}\) jika \(\lim f=7,\;\lim g=9\)
8. \(\lim_{x\to2}\frac{f(x)^2}{g(x)}\) jika \(\lim f=3,\;\lim g=9\)
9. \(\lim_{x\to1}[f(x)-g(x)]^2\) jika \(\lim f=5,\;\lim g=2\)
10. \(\lim_{x\to3}\frac{f(x)+1}{g(x)-1}\) jika \(\lim f=5,\;\lim g=3\)
11. \(\lim_{x\to1}\frac{f(x)^2-4}{f(x)-2}\) jika \(\lim f=2\)
12. \(\lim_{x\to0}\frac{f(x)^3-g(x)^3}{f(x)-g(x)}\) jika \(\lim f=1,\;\lim g=1\)
13. \(\lim_{x\to2}\left(\frac{f(x)}{g(x)}\right)^3\) jika \(\lim f=6,\;\lim g=2\)
14. \(\lim_{x\to1}\frac{\sqrt{f(x)}+\sqrt{g(x)}}{f(x)+g(x)}\) jika \(\lim f=4,\;\lim g=9\)
15. \(\lim_{x\to0}\frac{f(x)^2\cdot g(x)}{f(x)+g(x)^2}\) jika \(\lim f=3,\;\lim g=2\)
3. Strategi Menentukan Solusi Limit di Satu Titik
Langkah-langkah:
- Substitusi langsung — coba masukkan \(x=c\). Jika hasilnya terdefinisi, itulah limitnya.
- Jika bentuk \(\frac{0}{0}\) — faktorkan pembilang dan penyebut, lalu sederhanakan.
- Jika ada akar (bentuk irasional) — kalikan dengan sekawan (konjugat).
- Bentuk pangkat tinggi — gunakan pemfaktoran selisih/jumlah pangkat.
Contoh Soal
Mudah1. \(\displaystyle\lim_{x\to3}(x^2-2x+1)\)
Substitusi: \(9-6+1=4\)
2. \(\displaystyle\lim_{x\to-1}(x^3+x)\)
\((-1)^3+(-1)=-2\)
3. \(\displaystyle\lim_{x\to2}\frac{x+1}{x-1}\)
Substitusi: \(\frac{3}{1}=3\)
4. \(\displaystyle\lim_{x\to0}(x^2+4x+4)\)
\(0+0+4=4\)
5. \(\displaystyle\lim_{x\to1}\sqrt{x+3}\)
\(\sqrt{4}=2\)
6. \(\displaystyle\lim_{x\to2}\frac{x^2-4}{x^2-x-2}\)
Bentuk \(\frac{0}{0}\). Faktorkan:
\(\frac{(x-2)(x+2)}{(x-2)(x+1)}=\frac{x+2}{x+1}\)
Substitusi: \(\frac{4}{3}\)
7. \(\displaystyle\lim_{x\to-1}\frac{x^3+1}{x+1}\)
\(\frac{(x+1)(x^2-x+1)}{x+1}=x^2-x+1\)
Substitusi: \(1+1+1=3\)
8. \(\displaystyle\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}\)
\(\frac{(x-1)(x-2)}{(x-1)(x+1)}=\frac{x-2}{x+1}\)
Substitusi: \(\frac{-1}{2}\)
9. \(\displaystyle\lim_{x\to3}\frac{x^2-5x+6}{x^2-9}\)
\(\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3}\)
Substitusi: \(\frac{1}{6}\)
10. \(\displaystyle\lim_{x\to-2}\frac{x^3+8}{x+2}\)
\(\frac{(x+2)(x^2-2x+4)}{x+2}=x^2-2x+4\)
Substitusi: \(4+4+4=12\)
11. \(\displaystyle\lim_{x\to1}\frac{\sqrt{x+3}-2}{x-1}\)
Kalikan sekawan: \(\frac{(x+3)-4}{(x-1)(\sqrt{x+3}+2)}=\frac{x-1}{(x-1)(\sqrt{x+3}+2)}=\frac{1}{\sqrt{x+3}+2}\)
Substitusi: \(\frac{1}{4}\)
12. \(\displaystyle\lim_{x\to2}\frac{\sqrt{2x}-\sqrt{6-x}}{x-2}\)
Kalikan sekawan: \(\frac{2x-(6-x)}{(x-2)(\sqrt{2x}+\sqrt{6-x})}=\frac{3(x-2)}{(x-2)(\sqrt{2x}+\sqrt{6-x})}=\frac{3}{\sqrt{2x}+\sqrt{6-x}}\)
Substitusi: \(\frac{3}{2+2}=\frac{3}{4}\)
13. \(\displaystyle\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)
Kalikan sekawan: \(\frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}=\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}\)
Substitusi: \(\frac{2}{2}=1\)
14. \(\displaystyle\lim_{x\to4}\frac{x^2-16}{\sqrt{x}-2}\)
\(\frac{(x-4)(x+4)}{\sqrt{x}-2}\). Tulis \(x-4=(\sqrt{x}-2)(\sqrt{x}+2)\)
\(=(\sqrt{x}+2)(x+4)\). Substitusi: \((2+2)(8)=32\)
15. \(\displaystyle\lim_{x\to1}\frac{x^4-1}{x^3-1}\)
\(\frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)}=\frac{x^3+x^2+x+1}{x^2+x+1}\)
Substitusi: \(\frac{4}{3}\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x\to2}(x^2+x-1)\)
2. \(\displaystyle\lim_{x\to-3}(2x+5)\)
3. \(\displaystyle\lim_{x\to4}\frac{x}{x+1}\)
4. \(\displaystyle\lim_{x\to1}\sqrt{2x+7}\)
5. \(\displaystyle\lim_{x\to0}(x^3+2x^2+1)\)
6. \(\displaystyle\lim_{x\to5}\frac{x^2-25}{x^2-4x-5}\)
7. \(\displaystyle\lim_{x\to-1}\frac{x^2+3x+2}{x^2-1}\)
8. \(\displaystyle\lim_{x\to3}\frac{x^3-27}{x-3}\)
9. \(\displaystyle\lim_{x\to2}\frac{x^2-4}{x^3-8}\)
10. \(\displaystyle\lim_{x\to1}\frac{x^4-x}{x-1}\)
11. \(\displaystyle\lim_{x\to3}\frac{\sqrt{x+1}-2}{x-3}\)
12. \(\displaystyle\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}\)
13. \(\displaystyle\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}\)
14. \(\displaystyle\lim_{x\to1}\frac{x^5-1}{x^2-1}\)
15. \(\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-\sqrt{4-x}}{x}\)
4. Pengertian Limit di Tak Berhingga
Limit fungsi \(f(x)\) saat \(x\to\infty\) (atau \(x\to -\infty\)) menggambarkan perilaku fungsi ketika \(x\) membesar tanpa batas.
Artinya: semakin besar \(x\), nilai \(f(x)\) semakin mendekati \(L\).
Fakta penting:
- \(\displaystyle\lim_{x\to\infty}\frac{1}{x}=0\)
- \(\displaystyle\lim_{x\to\infty}\frac{1}{x^n}=0\) untuk \(n>0\)
- Untuk fungsi rasional, bandingkan derajat pembilang dan penyebut.
Contoh Soal
Mudah1. \(\displaystyle\lim_{x\to\infty}\frac{3}{x}\)
\(\frac{3}{\infty}=0\)
2. \(\displaystyle\lim_{x\to\infty}\frac{5}{x^2}\)
\(0\)
3. \(\displaystyle\lim_{x\to\infty}\frac{2x}{x}\)
Sederhanakan: \(2\)
4. \(\displaystyle\lim_{x\to\infty}(7-\frac{1}{x})\)
\(7-0=7\)
5. \(\displaystyle\lim_{x\to\infty}\frac{4x}{2x}\)
\(2\)
6. \(\displaystyle\lim_{x\to\infty}\frac{3x^2+1}{x^2-2}\)
Bagi pembilang dan penyebut dengan \(x^2\): \(\frac{3+\frac{1}{x^2}}{1-\frac{2}{x^2}}\to\frac{3}{1}=3\)
7. \(\displaystyle\lim_{x\to\infty}\frac{2x+3}{5x-1}\)
Bagi dengan \(x\): \(\frac{2}{5}\)
8. \(\displaystyle\lim_{x\to\infty}\frac{x^2+x}{3x^2}\)
Bagi \(x^2\): \(\frac{1+\frac{1}{x}}{3}\to\frac{1}{3}\)
9. \(\displaystyle\lim_{x\to\infty}\frac{x}{x^2+1}\)
Derajat pembilang < penyebut → 0
10. \(\displaystyle\lim_{x\to\infty}\frac{4x^3}{2x^3+x}\)
Bagi \(x^3\): \(\frac{4}{2}=2\)
11. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+4x}-x)\)
Kalikan sekawan: \(\frac{(x^2+4x)-x^2}{\sqrt{x^2+4x}+x}=\frac{4x}{\sqrt{x^2+4x}+x}\)
Bagi \(x\): \(\frac{4}{\sqrt{1+\frac{4}{x}}+1}\to\frac{4}{2}=2\)
12. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+6x+1}-x)\)
Sekawan: \(\frac{6x+1}{\sqrt{x^2+6x+1}+x}\). Bagi \(x\): \(\frac{6}{2}=3\)
13. \(\displaystyle\lim_{x\to\infty}(\sqrt{4x^2+x}-2x)\)
Sekawan: \(\frac{4x^2+x-4x^2}{\sqrt{4x^2+x}+2x}=\frac{x}{\sqrt{4x^2+x}+2x}\)
Bagi \(x\): \(\frac{1}{\sqrt{4}+2}=\frac{1}{4}\)
14. \(\displaystyle\lim_{x\to\infty}\frac{(2x+1)^3}{x^3+x}\)
Koefisien pangkat tertinggi pembilang: \(2^3=8\). Penyebut: 1.
Jadi = 8
15. \(\displaystyle\lim_{x\to\infty}(\sqrt{9x^2+3x}-3x)\)
Sekawan: \(\frac{3x}{\sqrt{9x^2+3x}+3x}\). Bagi \(x\): \(\frac{3}{3+3}=\frac{1}{2}\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x\to\infty}\frac{7}{x^3}\)
2. \(\displaystyle\lim_{x\to\infty}\frac{6x}{3x}\)
3. \(\displaystyle\lim_{x\to\infty}(2+\frac{1}{x^2})\)
4. \(\displaystyle\lim_{x\to\infty}\frac{10}{x}\)
5. \(\displaystyle\lim_{x\to\infty}\frac{x}{x}\)
6. \(\displaystyle\lim_{x\to\infty}\frac{5x^2-1}{2x^2+3}\)
7. \(\displaystyle\lim_{x\to\infty}\frac{x+2}{3x-1}\)
8. \(\displaystyle\lim_{x\to\infty}\frac{x^3+1}{2x^3-x}\)
9. \(\displaystyle\lim_{x\to\infty}\frac{2x}{x^2+1}\)
10. \(\displaystyle\lim_{x\to\infty}\frac{3x^2+x}{6x^2}\)
11. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+2x}-x)\)
12. \(\displaystyle\lim_{x\to\infty}(\sqrt{4x^2+8x}-2x)\)
13. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+10x+1}-x)\)
14. \(\displaystyle\lim_{x\to\infty}\frac{(x+3)^2}{x^2-1}\)
15. \(\displaystyle\lim_{x\to\infty}(\sqrt{9x^2+x}-3x)\)
5. Strategi Menentukan Solusi Limit di Tak Berhingga
Untuk fungsi rasional \(\frac{P(x)}{Q(x)}\):
- Jika derajat \(P\) < derajat \(Q\) → limit = 0
- Jika derajat \(P\) = derajat \(Q\) → limit = rasio koefisien tertinggi
- Jika derajat \(P\) > derajat \(Q\) → limit = ∞ (tidak hingga)
Untuk bentuk \(\sqrt{ax^2+bx+c}-dx\):
Kalikan dengan sekawan, lalu bagi pembilang dan penyebut dengan \(x\).
Rumus cepat: jika \(\sqrt{a}=d\), maka limit = \(\dfrac{b}{2\sqrt{a}}\)
Contoh Soal
Mudah1. \(\displaystyle\lim_{x\to\infty}\frac{2x}{x+1}\)
Derajat sama. Rasio koefisien: \(\frac{2}{1}=2\)
2. \(\displaystyle\lim_{x\to\infty}\frac{3}{x+5}\)
Derajat pembilang < penyebut → 0
3. \(\displaystyle\lim_{x\to\infty}\frac{x^2}{x^2+1}\)
Rasio: \(\frac{1}{1}=1\)
4. \(\displaystyle\lim_{x\to\infty}\frac{5x}{10x}\)
\(\frac{5}{10}=\frac{1}{2}\)
5. \(\displaystyle\lim_{x\to\infty}\frac{1}{2x+3}\)
0
6. \(\displaystyle\lim_{x\to\infty}\frac{4x^2-x+2}{2x^2+3x}\)
Rasio koefisien \(x^2\): \(\frac{4}{2}=2\)
7. \(\displaystyle\lim_{x\to\infty}\frac{x^3+2x}{3x^3-1}\)
\(\frac{1}{3}\)
8. \(\displaystyle\lim_{x\to\infty}\frac{x^2+1}{x^3}\)
Derajat pembilang < penyebut → 0
9. \(\displaystyle\lim_{x\to\infty}\frac{(x+1)(x-2)}{x^2+5}\)
Ekspansi pembilang: \(x^2-x-2\). Rasio: \(\frac{1}{1}=1\)
10. \(\displaystyle\lim_{x\to\infty}\frac{6x^2}{(2x+1)(x-3)}\)
Penyebut: \(2x^2-5x-3\). Rasio: \(\frac{6}{2}=3\)
11. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+8x}-x)\)
Rumus cepat: \(a=1, b=8\). Limit \(=\frac{8}{2(1)}=4\)
12. \(\displaystyle\lim_{x\to\infty}(\sqrt{4x^2+12x+1}-2x)\)
\(a=4,b=12,d=2=\sqrt4\). Limit \(=\frac{12}{2\cdot2}=3\)
13. \(\displaystyle\lim_{x\to\infty}\frac{(3x+1)^2-(x-2)^2}{2x^2}\)
Pembilang: \(9x^2+6x+1-x^2+4x-4=8x^2+10x-3\)
Rasio: \(\frac{8}{2}=4\)
14. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+5x+1}-\sqrt{x^2+x})\)
Sekawan: \(\frac{(x^2+5x+1)-(x^2+x)}{\sqrt{x^2+5x+1}+\sqrt{x^2+x}}=\frac{4x+1}{\sqrt{x^2+5x+1}+\sqrt{x^2+x}}\)
Bagi \(x\): \(\frac{4}{1+1}=2\)
15. \(\displaystyle\lim_{x\to\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}\)
Jadi: \(\frac{x}{\sqrt{x^2+1}+x}\). Bagi \(x\): \(\frac{1}{1+1}=\frac{1}{2}\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x\to\infty}\frac{7x}{x-2}\)
2. \(\displaystyle\lim_{x\to\infty}\frac{4}{3x}\)
3. \(\displaystyle\lim_{x\to\infty}\frac{x}{2x}\)
4. \(\displaystyle\lim_{x\to\infty}\frac{2x+1}{x}\)
5. \(\displaystyle\lim_{x\to\infty}\frac{3x^2}{x^2+x}\)
6. \(\displaystyle\lim_{x\to\infty}\frac{2x^3-x}{5x^3+2}\)
7. \(\displaystyle\lim_{x\to\infty}\frac{x^2-3x}{4x^2+1}\)
8. \(\displaystyle\lim_{x\to\infty}\frac{(x+2)^2}{x^2}\)
9. \(\displaystyle\lim_{x\to\infty}\frac{x^2}{x^3-x}\)
10. \(\displaystyle\lim_{x\to\infty}\frac{2x^4+1}{x^4+x^2}\)
11. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+12x}-x)\)
12. \(\displaystyle\lim_{x\to\infty}(\sqrt{9x^2+6x}-3x)\)
13. \(\displaystyle\lim_{x\to\infty}(\sqrt{x^2+3x}-\sqrt{x^2+x})\)
14. \(\displaystyle\lim_{x\to\infty}\frac{(2x-1)^3}{4x^3+x}\)
15. \(\displaystyle\lim_{x\to\infty}x(\sqrt{x^2+4}-x)\)
Limit Fungsi Trigonometri
Limit Fungsi Trigonometri
Materi lengkap meliputi Limit Fungsi Sinus, Kosinus, dan Tangen — disertai contoh soal & latihan
Daftar Isi
1 Limit Fungsi Sinus
📘 Materi
Limit fungsi sinus merupakan salah satu limit paling fundamental dalam kalkulus. Rumus dasar yang harus dikuasai:
Rumus Dasar Limit Sinus:
\[\lim_{x \to 0} \frac{\sin x}{x} = 1\]
\[\lim_{x \to 0} \frac{x}{\sin x} = 1\]
\[\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}\]
\[\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}\]
Penjelasan:
- Rumus ini hanya berlaku saat \(x \to 0\).
- Sudut \(x\) harus dalam radian.
- Bentuk \(\frac{\sin ax}{bx}\) dapat diubah menjadi \(\frac{a}{b} \cdot \frac{\sin ax}{ax}\), lalu gunakan rumus dasar.
- Jika bentuk limit bukan \(\frac{0}{0}\), substitusi langsung bisa digunakan.
Tabel Nilai Pendekatan \(\frac{\sin x}{x}\) saat \(x \to 0\):
| \(x\) (rad) | \(\sin x\) | \(\frac{\sin x}{x}\) |
|---|---|---|
| 0.5 | 0.4794 | 0.9589 |
| 0.1 | 0.0998 | 0.9983 |
| 0.01 | 0.00999 | 0.99998 |
| 0.001 | 0.000999 | 0.9999998 |
Semakin kecil \(x\), nilainya semakin mendekati 1.
📝 Contoh Soal — Limit Sinus
Klik soal untuk melihat/menyembunyikan pembahasan.
MUDAH
Pembahasan:
\[\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{3 \cdot \sin 3x}{3x} = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = \boxed{3}\]
Pembahasan:
\[\lim_{x \to 0} \frac{\sin 5x}{\sin 2x} = \frac{5}{2} = \boxed{\frac{5}{2}}\]
Menggunakan rumus \(\lim_{x\to 0}\frac{\sin ax}{\sin bx}=\frac{a}{b}\).
Pembahasan:
\[\lim_{x \to 0} \frac{4x}{\sin 2x} = \frac{4}{2} \cdot \lim_{x\to 0}\frac{2x}{\sin 2x} = 2 \cdot 1 = \boxed{2}\]
Pembahasan:
\[\lim_{x \to 0} \frac{\sin x}{3x} = \frac{1}{3}\cdot\lim_{x\to 0}\frac{\sin x}{x} = \frac{1}{3}\cdot 1 = \boxed{\frac{1}{3}}\]
Pembahasan:
\[\lim_{x \to 0} \frac{\sin 4x}{4x} = 1\]
Langsung menggunakan rumus dasar \(\lim_{x\to 0}\frac{\sin u}{u}=1\) dengan \(u=4x\). Jawaban: \(\boxed{1}\)
SEDANG
Pembahasan:
\[\lim_{x \to 0} \frac{\sin 3x – \sin x}{x} = \lim_{x\to 0}\frac{\sin 3x}{x} – \lim_{x\to 0}\frac{\sin x}{x} = 3 – 1 = \boxed{2}\]
Pembahasan:
\[\lim_{x \to 0} \frac{\sin^2 2x}{x^2} = \lim_{x\to 0}\left(\frac{\sin 2x}{x}\right)^2 = \left(\frac{2}{1}\right)^2 = \boxed{4}\]
Pembahasan:
\[\frac{x \sin 2x}{\sin^2 3x} = \frac{x}{\sin 3x}\cdot\frac{\sin 2x}{\sin 3x}\]
\[= \frac{1}{3}\cdot\frac{3x}{\sin 3x}\cdot\frac{2}{3}\cdot\frac{\sin 2x}{2x}\cdot\frac{3x}{\sin 3x}\]
Saat \(x\to 0\): \(= \frac{1}{3}\cdot 1 \cdot \frac{2}{3}\cdot 1 \cdot 1 = \boxed{\frac{2}{9}}\)
Pembahasan:
\[= \lim_{x\to 0}\frac{\sin 5x}{\sin 4x} + \lim_{x\to 0}\frac{\sin 3x}{\sin 4x} = \frac{5}{4}+\frac{3}{4} = \boxed{2}\]
Pembahasan:
\[= \lim_{x\to 0}\frac{\sin 2x}{x}\cdot\lim_{x\to 0}\frac{\sin 3x}{x} = 2 \cdot 3 = \boxed{6}\]
SULIT
Pembahasan:
Gunakan identitas \(1-\cos 2x = 2\sin^2 x\):
\[\lim_{x\to 0}\frac{2\sin^2 x}{x\sin x} = \lim_{x\to 0}\frac{2\sin x}{x} = 2\cdot 1 = \boxed{2}\]
Pembahasan:
Substitusi \(x = \frac{\pi}{6}+h\), saat \(h\to 0\):
\(2\sin x – 1 = 2\sin(\tfrac{\pi}{6}+h)-1 = 2(\tfrac{1}{2}\cos h + \tfrac{\sqrt{3}}{2}\sin h)-1 = \cos h – 1 + \sqrt{3}\sin h\)
\(\sin 3x = \sin(\tfrac{\pi}{2}+3h) = \cos 3h\)
Saat \(h\to 0\): pembilang \(\to 0 + \sqrt{3}\cdot 0 = 0\) tapi \(\cos 3h \to 1\).
Lebih teliti: \(\frac{\cos h -1+\sqrt{3}\sin h}{\cos 3h}\). Saat \(h\to 0\), penyebut \(\to 1\), jadi kita evaluasi pembilang. \(\cos h – 1 \approx -\frac{h^2}{2}\), \(\sqrt{3}\sin h \approx \sqrt{3}h\).
Pembilang \(\approx \sqrt{3}h\), penyebut \(\to 1\). Jadi limit \(= 0\)?
Periksa: substitusi langsung \(x=\frac{\pi}{6}\): pembilang \(= 2\cdot\frac{1}{2}-1=0\), penyebut \(=\sin\frac{\pi}{2}=1\).
Bukan bentuk \(\frac{0}{0}\), maka substitusi langsung:
\[\frac{2\sin\frac{\pi}{6}-1}{\sin\frac{\pi}{2}} = \frac{0}{1} = \boxed{0}\]
Pembahasan:
Gunakan deret Taylor: \(\sin x = x – \frac{x^3}{6}+\cdots\), \(\cos x = 1-\frac{x^2}{2}+\cdots\)
\[\sin x – x\cos x = \left(x-\frac{x^3}{6}+\cdots\right) – x\left(1-\frac{x^2}{2}+\cdots\right)\]
\[= x – \frac{x^3}{6} – x + \frac{x^3}{2} + \cdots = \frac{x^3}{3}+\cdots\]
\[\lim_{x\to 0}\frac{\frac{x^3}{3}}{x^3} = \boxed{\frac{1}{3}}\]
Pembahasan:
Deret Taylor: \(x – \sin x = x – (x – \frac{x^3}{6}+\cdots) = \frac{x^3}{6}+\cdots\)
\(x^2\sin x = x^2(x – \frac{x^3}{6}+\cdots) = x^3 – \frac{x^5}{6}+\cdots\)
\[\lim_{x\to 0}\frac{\frac{x^3}{6}}{x^3} = \boxed{\frac{1}{6}}\]
Pembahasan:
Misalkan \(u = \sin x\), saat \(x\to 0\) maka \(u\to 0\):
\[\lim_{x\to 0}\frac{\sin(\sin x)}{x} = \lim_{x\to 0}\frac{\sin(\sin x)}{\sin x}\cdot\frac{\sin x}{x}\]
\[= \lim_{u\to 0}\frac{\sin u}{u}\cdot\lim_{x\to 0}\frac{\sin x}{x} = 1 \cdot 1 = \boxed{1}\]
✏️ Latihan Soal — Limit Sinus
MUDAH
- \(\displaystyle\lim_{x \to 0} \frac{\sin 7x}{x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 2x}{\sin 5x}\)
- \(\displaystyle\lim_{x \to 0} \frac{3x}{\sin 6x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 10x}{2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 3x}{\sin 3x}\)
SEDANG
- \(\displaystyle\lim_{x \to 0} \frac{\sin^2 3x}{x \sin 2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 4x + \sin 6x}{2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{x^2}{\sin 2x \cdot \sin 3x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 2x – \sin x}{\sin 3x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 5x \cdot \sin x}{x^2}\)
SULIT
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 4x}{x\sin 2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin x – x}{x^3}\)
- \(\displaystyle\lim_{x \to 0} \frac{x\sin x}{1-\cos x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin(x^2)}{x\sin x}\)
- \(\displaystyle\lim_{x \to \pi} \frac{\sin x}{x – \pi}\)
2 Limit Fungsi Kosinus
📘 Materi
Limit fungsi kosinus sering melibatkan bentuk \(1-\cos x\). Identitas trigonometri sangat penting di sini.
Rumus Dasar Limit Kosinus:
\[\lim_{x \to 0} \cos x = 1\]
\[\lim_{x \to 0} \frac{1 – \cos x}{x} = 0\]
\[\lim_{x \to 0} \frac{1 – \cos x}{x^2} = \frac{1}{2}\]
\[\lim_{x \to 0} \frac{1 – \cos ax}{x^2} = \frac{a^2}{2}\]
Identitas Penting:
\[1 – \cos x = 2\sin^2\frac{x}{2}\]
\[1 – \cos 2x = 2\sin^2 x\]
\[\cos^2 x = \frac{1+\cos 2x}{2}\]
Strategi umum: ubah \(1-\cos x\) menjadi \(2\sin^2\frac{x}{2}\) lalu gunakan rumus limit sinus.
📝 Contoh Soal — Limit Kosinus
Klik soal untuk melihat/menyembunyikan pembahasan.
MUDAH
Pembahasan:
\[\frac{1-\cos x}{x^2} = \frac{2\sin^2\frac{x}{2}}{x^2} = \frac{2\sin^2\frac{x}{2}}{4\cdot\frac{x^2}{4}} = \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\]
Saat \(x\to 0\): \(= \frac{1}{2}\cdot 1^2 = \boxed{\frac{1}{2}}\)
Pembahasan:
\[\frac{1-\cos x}{x} = \frac{2\sin^2\frac{x}{2}}{x} = \sin\frac{x}{2}\cdot\frac{\sin\frac{x}{2}}{\frac{x}{2}}\cdot\frac{1}{1}\]
Saat \(x\to 0\): \(\sin\frac{x}{2}\to 0\) dan \(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\to 1\), jadi hasilnya \(= 0\cdot 1 = \boxed{0}\)
Pembahasan:
Menggunakan rumus \(\lim_{x\to 0}\frac{1-\cos ax}{x^2}=\frac{a^2}{2}\) dengan \(a=4\):
\[= \frac{4^2}{2} = \frac{16}{2} = \boxed{8}\]
Pembahasan:
\[= \frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos 3x}{x^2}} = \frac{\frac{4}{2}}{\frac{9}{2}} = \frac{4}{9} = \boxed{\frac{4}{9}}\]
Pembahasan:
\[= \frac{1-\cos 6x}{x^2}\cdot\frac{x}{\sin 3x}\cdot\frac{x}{1}\cdot\frac{1}{1}\]
Koreksi: \(\frac{1-\cos 6x}{\sin 3x \cdot x} = \frac{1-\cos 6x}{x^2}\cdot\frac{x}{\sin 3x}\)
\(= \frac{36}{2}\cdot\frac{1}{3} = 18\cdot\frac{1}{3} = \boxed{6}\)
SEDANG
Pembahasan:
\(1-\cos^2 x = \sin^2 x\), maka:
\[\lim_{x\to 0}\frac{\sin^2 x}{x^2} = \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 = 1^2 = \boxed{1}\]
Pembahasan:
\[\cos 2x – \cos 4x = (1-(1-\cos 2x)) – (1-(1-\cos 4x))\]
\[= -(1-\cos 2x)+(1-\cos 4x) = (1-\cos 4x)-(1-\cos 2x)\]
\[\frac{(1-\cos 4x)-(1-\cos 2x)}{x^2} = \frac{16}{2}-\frac{4}{2} = 8-2 = \boxed{6}\]
Pembahasan:
\(1-\cos 2x = 2\sin^2 x\), maka:
\[\frac{2\sin^2 x}{\sin^2 x} = \boxed{2}\]
Pembahasan:
\[= \frac{1-\cos x}{x^2}\cdot\frac{x^2}{\sin^2 2x} = \frac{1}{2}\cdot\frac{1}{4} = \boxed{\frac{1}{8}}\]
Karena \(\frac{x^2}{\sin^2 2x}=\left(\frac{x}{\sin 2x}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
Pembahasan:
\(1-\cos 2x = 2\sin^2 x\):
\[\frac{x\cdot 2\sin^2 x}{\sin^3 x} = \frac{2x}{\sin x} = 2\cdot\frac{x}{\sin x}\]
Saat \(x\to 0\): \(= 2\cdot 1 = \boxed{2}\)
SULIT
Pembahasan:
Gunakan rumus jumlah ke kali: \(\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\)
\[\cos x – \cos 3x = -2\sin 2x\sin(-x) = 2\sin 2x\sin x\]
\[\frac{2\sin 2x\sin x}{x\sin x} = \frac{2\sin 2x}{x} = 2\cdot\frac{\sin 2x}{x} = 2\cdot 2 = \boxed{4}\]
Pembahasan:
Misalkan \(h = x – \frac{\pi}{2}\), saat \(x\to\frac{\pi}{2}\) maka \(h\to 0\):
\(\cos x = \cos(\frac{\pi}{2}+h) = -\sin h\)
\[\lim_{h\to 0}\frac{-\sin h}{h} = -1 = \boxed{-1}\]
Pembahasan:
Tambah dan kurangi:
\[1-\cos x\cos 2x = (1-\cos x)+\cos x(1-\cos 2x)\]
\[\frac{(1-\cos x)}{x^2}+\frac{\cos x(1-\cos 2x)}{x^2}\]
Saat \(x\to 0\): \(= \frac{1}{2}+1\cdot\frac{4}{2} = \frac{1}{2}+2 = \boxed{\frac{5}{2}}\)
Pembahasan:
\(1+\cos 2x = 2\cos^2 x\), maka \(\sqrt{1+\cos 2x}=\sqrt{2}|\cos x|=\sqrt{2}\cos x\) (dekat 0).
\[\frac{\sqrt{2}\cos x – \sqrt{2}}{x^2} = \sqrt{2}\cdot\frac{\cos x – 1}{x^2} = \sqrt{2}\cdot\left(-\frac{1}{2}\right) = \boxed{-\frac{\sqrt{2}}{2}}\]
Pembahasan:
Tuliskan: \(1-\cos x\cos 2x\cos 3x\)
\(= (1-\cos x)+\cos x(1-\cos 2x)+\cos x\cos 2x(1-\cos 3x)\)
Bagi \(x^2\) dan ambil limit:
\[= \frac{1}{2}+1\cdot\frac{4}{2}+1\cdot 1\cdot\frac{9}{2} = \frac{1}{2}+2+\frac{9}{2} = \frac{1+4+9}{2} = \boxed{7}\]
✏️ Latihan Soal — Limit Kosinus
MUDAH
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 3x}{x^2}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 5x}{x}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos x}{\sin x}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 2x}{4x^2}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos x}{x\sin x}\)
SEDANG
- \(\displaystyle\lim_{x \to 0} \frac{\cos 3x – \cos 5x}{x^2}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 2x}{x\sin 3x}\)
- \(\displaystyle\lim_{x \to 0} \frac{(1-\cos x)^2}{x^4}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 4x}{\sin^2 2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\cos x – \cos 5x}{\sin^2 x}\)
SULIT
- \(\displaystyle\lim_{x \to 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{x^2}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos x\cos 2x\cos 3x\cos 4x}{x^2}\)
- \(\displaystyle\lim_{x \to \frac{\pi}{3}} \frac{1-2\cos x}{\pi – 3x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\cos ax – \cos bx}{x^2}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos(\sin x)}{x^2}\)
3 Limit Fungsi Tangen
📘 Materi
Limit fungsi tangen erat kaitannya dengan limit sinus dan kosinus karena \(\tan x = \frac{\sin x}{\cos x}\).
Rumus Dasar Limit Tangen:
\[\lim_{x \to 0} \frac{\tan x}{x} = 1\]
\[\lim_{x \to 0} \frac{x}{\tan x} = 1\]
\[\lim_{x \to 0} \frac{\tan ax}{bx} = \frac{a}{b}\]
\[\lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac{a}{b}\]
\[\lim_{x \to 0} \frac{\sin ax}{\tan bx} = \frac{a}{b}\]
Penjelasan:
- Semua rumus di atas berlaku saat \(x\to 0\).
- Kunci utama: \(\tan x = \frac{\sin x}{\cos x}\), dan \(\cos 0 = 1\), sehingga di dekat 0, \(\tan x \approx \sin x\).
- Untuk soal yang lebih kompleks, ubah \(\tan\) ke bentuk \(\frac{\sin}{\cos}\) lalu gunakan rumus limit sinus.
Tabel Perbandingan \(\frac{\tan x}{x}\) vs \(\frac{\sin x}{x}\):
| \(x\) | \(\frac{\sin x}{x}\) | \(\frac{\tan x}{x}\) |
|---|---|---|
| 0.5 | 0.9589 | 1.0926 |
| 0.1 | 0.9983 | 1.0033 |
| 0.01 | 0.99998 | 1.00003 |
Keduanya mendekati 1 saat \(x\to 0\), tetapi \(\frac{\tan x}{x}\) mendekati dari atas.
📝 Contoh Soal — Limit Tangen
Klik soal untuk melihat/menyembunyikan pembahasan.
MUDAH
Pembahasan:
\[\lim_{x\to 0}\frac{\tan 3x}{x} = 3\cdot\lim_{x\to 0}\frac{\tan 3x}{3x} = 3\cdot 1 = \boxed{3}\]
Pembahasan:
\[\lim_{x\to 0}\frac{\tan 2x}{\tan 5x} = \frac{2}{5} = \boxed{\frac{2}{5}}\]
Pembahasan:
\[\lim_{x\to 0}\frac{\sin 3x}{\tan 2x} = \frac{3}{2} = \boxed{\frac{3}{2}}\]
Pembahasan:
\[\frac{6x}{\tan 3x} = \frac{6}{3}\cdot\frac{3x}{\tan 3x} = 2\cdot 1 = \boxed{2}\]
Pembahasan:
\[\frac{\tan 4x}{\sin 4x} = \frac{\sin 4x}{\cos 4x \cdot \sin 4x} = \frac{1}{\cos 4x}\]
Saat \(x\to 0\): \(\frac{1}{\cos 0} = \boxed{1}\)
SEDANG
Pembahasan:
\[\tan 2x – \sin 2x = \frac{\sin 2x}{\cos 2x}-\sin 2x = \sin 2x\left(\frac{1}{\cos 2x}-1\right) = \sin 2x\cdot\frac{1-\cos 2x}{\cos 2x}\]
\[= \frac{\sin 2x \cdot 2\sin^2 x}{\cos 2x}\]
\[\frac{\sin 2x \cdot 2\sin^2 x}{x^3\cos 2x} = \frac{\sin 2x}{x}\cdot\frac{2\sin^2 x}{x^2}\cdot\frac{1}{\cos 2x} = 2\cdot 2\cdot 1 = \boxed{4}\]
Pembahasan:
\[= \frac{\tan 3x}{x}\cdot\frac{\tan 3x}{\sin 2x} = 3\cdot\frac{3}{2} = \boxed{\frac{9}{2}}\]
Pembahasan:
\(1-\cos 2x = 2\sin^2 x\):
\[\frac{2\sin^2 x}{\tan^2 x} = \frac{2\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}} = 2\cos^2 x\]
Saat \(x\to 0\): \(= 2\cdot 1 = \boxed{2}\)
Pembahasan:
\[= \frac{\sin 4x}{3x}+\frac{\tan 2x}{3x} = \frac{4}{3}+\frac{2}{3} = \boxed{2}\]
Pembahasan:
\(1-\cos x = 2\sin^2\frac{x}{2}\):
\[\frac{x\tan x}{2\sin^2\frac{x}{2}} = \frac{x}{\sin\frac{x}{2}}\cdot\frac{\tan x}{\sin\frac{x}{2}}\cdot\frac{1}{2}\]
Alternatif: \(\frac{x\tan x}{1-\cos x} = \frac{x}{1}\cdot\frac{\tan x}{x}\cdot\frac{x^2}{1-\cos x}\cdot\frac{1}{x}\)
Lebih simpel: \(= \frac{x\tan x}{x^2}\cdot\frac{x^2}{1-\cos x} = \frac{\tan x}{x}\cdot\frac{1}{\frac{1-\cos x}{x^2}} = \frac{1}{\frac{1}{2}} = \boxed{2}\)
SULIT
Pembahasan:
\[\tan x – 1 = \frac{\sin x – \cos x}{\cos x}\]
\[\frac{\tan x – 1}{\sin x – \cos x} = \frac{\sin x – \cos x}{\cos x(\sin x – \cos x)} = \frac{1}{\cos x}\]
Saat \(x\to\frac{\pi}{4}\): \(\frac{1}{\cos\frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \boxed{\sqrt{2}}\)
Pembahasan:
\[\tan x – \sin x = \frac{\sin x}{\cos x}-\sin x = \sin x\cdot\frac{1-\cos x}{\cos x}\]
\[\frac{\sin x(1-\cos x)}{x^3\cos x} = \frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}\cdot\frac{1}{\cos x} = 1\cdot\frac{1}{2}\cdot 1 = \boxed{\frac{1}{2}}\]
Pembahasan:
Ini adalah limit klasik yang terkenal sulit. Menggunakan ekspansi deret Taylor orde tinggi:
\(\sin x = x – \frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots\)
\(\tan x = x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\cdots\)
Setelah perhitungan panjang (substitusi dan ekspansi hingga orde 7):
\[\tan(\sin x)-\sin(\tan x) = -\frac{x^7}{30}+\cdots\]
\[\lim_{x\to 0}\frac{-\frac{x^7}{30}}{x^7} = \boxed{-\frac{1}{30}}\]
Pembahasan:
Faktorkan: \(a^3-b^3 = (a-b)(a^2+ab+b^2)\) dengan \(a=\tan 2x, b=\sin 2x\):
\(\tan 2x – \sin 2x = \sin 2x\cdot\frac{1-\cos 2x}{\cos 2x}\)
Saat \(x\to 0\): \(\tan^2 2x + \tan 2x\sin 2x + \sin^2 2x \approx (2x)^2+(2x)(2x)+(2x)^2 = 12x^2\)
\(\tan 2x – \sin 2x \approx 2x\cdot\frac{2x^2}{1}\cdot 2 = 4x^3\) (lebih tepatnya).
Dengan ekspansi: \(\tan 2x – \sin 2x \approx \frac{(2x)^3}{2}\cdot\frac{1}{1} = 4x^3\)
Maka: \(\frac{4x^3\cdot 12x^2}{x^5} = \frac{48x^5}{x^5} = \boxed{48}\)
Pembahasan:
Substitusi \(x = \frac{\pi}{4}+h\), \(h\to 0\):
\(\tan x = \tan(\frac{\pi}{4}+h) = \frac{1+\tan h}{1-\tan h}\)
\(1-\tan x = 1-\frac{1+\tan h}{1-\tan h} = \frac{-2\tan h}{1-\tan h}\)
\(\sqrt{2}\sin x = \sqrt{2}\sin(\frac{\pi}{4}+h) = \sqrt{2}(\frac{\sqrt{2}}{2}\cos h+\frac{\sqrt{2}}{2}\sin h) = \cos h+\sin h\)
\(1-\sqrt{2}\sin x = 1-\cos h – \sin h\)
Saat \(h\to 0\): pembilang \(\approx \frac{-2h}{1} = -2h\), penyebut \(\approx \frac{h^2}{2}-h = -h+\frac{h^2}{2}\approx -h\)
\[\frac{-2h}{-h} = \boxed{2}\]
✏️ Latihan Soal — Limit Tangen
MUDAH
- \(\displaystyle\lim_{x \to 0} \frac{\tan 5x}{x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\tan 3x}{\tan 7x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 2x}{\tan 6x}\)
- \(\displaystyle\lim_{x \to 0} \frac{8x}{\tan 2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\tan x}{\sin x}\)
SEDANG
- \(\displaystyle\lim_{x \to 0} \frac{\tan^2 2x}{\sin 3x \cdot x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\sin 3x + \tan 5x}{4x}\)
- \(\displaystyle\lim_{x \to 0} \frac{1-\cos 4x}{\tan^2 2x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\tan 3x – \sin 3x}{x^3}\)
- \(\displaystyle\lim_{x \to 0} \frac{x^2}{\tan 2x\cdot\sin 3x}\)
SULIT
- \(\displaystyle\lim_{x \to \frac{\pi}{4}} \frac{\tan x – 1}{x – \frac{\pi}{4}}\)
- \(\displaystyle\lim_{x \to 0} \frac{\tan x – x}{x – \sin x}\)
- \(\displaystyle\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}-2\sin x}{1-2\sin^2 x}\)
- \(\displaystyle\lim_{x \to 0} \frac{\tan(\tan x)-\tan x}{x^3}\)
- \(\displaystyle\lim_{x \to 0} \frac{e^{\tan x}-e^x}{x^3}\) (bonus: melibatkan eksponen)
Bentuk Tak Tentu Limit Fungsi
Bentuk Tak Tentu Limit Fungsi
Matematika Kelas XI/XII — Materi, Contoh Soal & Latihan
Pendahuluan
Dalam menghitung limit fungsi, kita sering menjumpai bentuk yang tidak dapat langsung ditentukan nilainya. Bentuk-bentuk ini disebut bentuk tak tentu (indeterminate form). Empat bentuk tak tentu utama yang akan dipelajari:
- \(\dfrac{0}{0}\)
- \(\dfrac{\infty}{\infty}\)
- \(0 \cdot \infty\)
- \(\infty – \infty\)
Untuk menyelesaikan bentuk tak tentu, kita perlu melakukan manipulasi aljabar seperti memfaktorkan, mengalikan sekawan, atau membagi pangkat tertinggi.
1. Bentuk Tak Tentu \(\dfrac{0}{0}\)
Materi
Bentuk \(\dfrac{0}{0}\) terjadi ketika substitusi langsung menghasilkan pembilang = 0 dan penyebut = 0. Strategi penyelesaian:
- Faktorisasi — faktorkan pembilang dan penyebut, lalu coret faktor yang sama.
- Mengalikan sekawan — jika terdapat bentuk akar.
- Substitusi variabel — untuk bentuk akar pangkat tertentu.
Contoh Soal
Mudah1. Hitung \(\displaystyle\lim_{x \to 2} \frac{x^2 – 4}{x – 2}\)
▶ Lihat PembahasanSubstitusi langsung: \(\frac{4-4}{2-2} = \frac{0}{0}\) → bentuk tak tentu.
Faktorisasi: \(\frac{(x-2)(x+2)}{x-2} = x+2\)
Maka: \(\displaystyle\lim_{x \to 2}(x+2) = 4\)
2. Hitung \(\displaystyle\lim_{x \to 3} \frac{x^2 – 9}{x – 3}\)
▶ Lihat PembahasanFaktorisasi: \(\frac{(x-3)(x+3)}{x-3} = x+3\)
\(\displaystyle\lim_{x \to 3}(x+3) = 6\)
3. Hitung \(\displaystyle\lim_{x \to 1} \frac{x^2 – 1}{x – 1}\)
▶ Lihat Pembahasan\(\frac{(x-1)(x+1)}{x-1} = x+1\)
\(\displaystyle\lim_{x \to 1}(x+1) = 2\)
4. Hitung \(\displaystyle\lim_{x \to 5} \frac{x^2 – 25}{x – 5}\)
▶ Lihat Pembahasan\(\frac{(x-5)(x+5)}{x-5} = x+5\)
\(\displaystyle\lim_{x \to 5}(x+5) = 10\)
5. Hitung \(\displaystyle\lim_{x \to -1} \frac{x^2 + 3x + 2}{x + 1}\)
▶ Lihat PembahasanFaktorisasi pembilang: \(x^2+3x+2 = (x+1)(x+2)\)
\(\frac{(x+1)(x+2)}{x+1} = x+2\)
\(\displaystyle\lim_{x \to -1}(x+2) = 1\)
6. Hitung \(\displaystyle\lim_{x \to 4} \frac{\sqrt{x} – 2}{x – 4}\)
▶ Lihat PembahasanKalikan sekawan: \(\frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}\)
\(\displaystyle\lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{2+2} = \frac{1}{4}\)
7. Hitung \(\displaystyle\lim_{x \to 1} \frac{x^3 – 1}{x – 1}\)
▶ Lihat PembahasanFaktorisasi: \(x^3 – 1 = (x-1)(x^2+x+1)\)
\(\frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1\)
\(\displaystyle\lim_{x \to 1}(1+1+1) = 3\)
8. Hitung \(\displaystyle\lim_{x \to 2} \frac{x^2 – 5x + 6}{x^2 – 4}\)
▶ Lihat PembahasanFaktorisasi: \(\frac{(x-2)(x-3)}{(x-2)(x+2)} = \frac{x-3}{x+2}\)
\(\displaystyle\lim_{x \to 2} \frac{2-3}{2+2} = \frac{-1}{4}\)
9. Hitung \(\displaystyle\lim_{x \to 0} \frac{\sqrt{1+x} – 1}{x}\)
▶ Lihat PembahasanKalikan sekawan: \(\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \frac{1+x-1}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}\)
\(\displaystyle\lim_{x \to 0} \frac{1}{\sqrt{1}+1} = \frac{1}{2}\)
10. Hitung \(\displaystyle\lim_{x \to 9} \frac{x – 9}{\sqrt{x} – 3}\)
▶ Lihat PembahasanKalikan sekawan penyebut: \(\frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} = \frac{(x-9)(\sqrt{x}+3)}{x-9} = \sqrt{x}+3\)
\(\displaystyle\lim_{x \to 9}(\sqrt{9}+3) = 3+3 = 6\)
11. Hitung \(\displaystyle\lim_{x \to 1} \frac{\sqrt{3x+1} – \sqrt{5x-1}}{x^2 – 1}\)
▶ Lihat PembahasanKalikan sekawan pembilang:
\(\frac{(\sqrt{3x+1}-\sqrt{5x-1})(\sqrt{3x+1}+\sqrt{5x-1})}{(x^2-1)(\sqrt{3x+1}+\sqrt{5x-1})} = \frac{(3x+1)-(5x-1)}{(x^2-1)(\sqrt{3x+1}+\sqrt{5x-1})}\)
\(= \frac{-2x+2}{(x-1)(x+1)(\sqrt{3x+1}+\sqrt{5x-1})} = \frac{-2(x-1)}{(x-1)(x+1)(\sqrt{3x+1}+\sqrt{5x-1})}\)
\(= \frac{-2}{(x+1)(\sqrt{3x+1}+\sqrt{5x-1})}\)
\(\displaystyle\lim_{x \to 1} = \frac{-2}{2(\sqrt{4}+\sqrt{4})} = \frac{-2}{2(2+2)} = \frac{-2}{8} = -\frac{1}{4}\)
12. Hitung \(\displaystyle\lim_{x \to 0} \frac{\sqrt{1+x} – \sqrt{1-x}}{x}\)
▶ Lihat PembahasanKalikan sekawan: \(\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})} = \frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}\)
\(= \frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})} = \frac{2}{\sqrt{1+x}+\sqrt{1-x}}\)
\(\displaystyle\lim_{x \to 0} = \frac{2}{1+1} = 1\)
13. Hitung \(\displaystyle\lim_{x \to 8} \frac{\sqrt[3]{x} – 2}{x – 8}\)
▶ Lihat PembahasanMisalkan \(u = \sqrt[3]{x}\), maka \(x = u^3\) dan saat \(x \to 8\), \(u \to 2\).
\(\frac{u – 2}{u^3 – 8} = \frac{u-2}{(u-2)(u^2+2u+4)} = \frac{1}{u^2+2u+4}\)
\(\displaystyle\lim_{u \to 2} \frac{1}{4+4+4} = \frac{1}{12}\)
14. Hitung \(\displaystyle\lim_{x \to 1} \frac{x^3 – 3x + 2}{x^3 – x^2 – x + 1}\)
▶ Lihat PembahasanPembilang: \(x^3-3x+2 = (x-1)^2(x+2)\)
Penyebut: \(x^3-x^2-x+1 = (x-1)^2(x+1)\)
\(\frac{(x-1)^2(x+2)}{(x-1)^2(x+1)} = \frac{x+2}{x+1}\)
\(\displaystyle\lim_{x \to 1} \frac{3}{2} = \frac{3}{2}\)
15. Hitung \(\displaystyle\lim_{x \to 0} \frac{\sqrt[3]{1+x} – 1}{x}\)
▶ Lihat PembahasanMisalkan \(u = \sqrt[3]{1+x}\), maka \(u^3 = 1+x\), \(x = u^3 – 1\), saat \(x\to 0\), \(u\to 1\).
\(\frac{u-1}{u^3-1} = \frac{u-1}{(u-1)(u^2+u+1)} = \frac{1}{u^2+u+1}\)
\(\displaystyle\lim_{u \to 1} \frac{1}{1+1+1} = \frac{1}{3}\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x \to 4} \frac{x^2 – 16}{x – 4}\)
2. \(\displaystyle\lim_{x \to -2} \frac{x^2 – 4}{x + 2}\)
3. \(\displaystyle\lim_{x \to 3} \frac{x^2 – 6x + 9}{x – 3}\)
4. \(\displaystyle\lim_{x \to 0} \frac{x^2 + 5x}{x}\)
5. \(\displaystyle\lim_{x \to 6} \frac{x^2 – 36}{x – 6}\)
6. \(\displaystyle\lim_{x \to 1} \frac{x^3 – x}{x – 1}\)
7. \(\displaystyle\lim_{x \to 16} \frac{\sqrt{x} – 4}{x – 16}\)
8. \(\displaystyle\lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4}\)
9. \(\displaystyle\lim_{x \to 0} \frac{\sqrt{4+x} – 2}{x}\)
10. \(\displaystyle\lim_{x \to 3} \frac{x^2 – 2x – 3}{x^2 – 9}\)
11. \(\displaystyle\lim_{x \to 4} \frac{\sqrt{2x+1} – 3}{x^2 – 16}\)
12. \(\displaystyle\lim_{x \to 27} \frac{\sqrt[3]{x} – 3}{x – 27}\)
13. \(\displaystyle\lim_{x \to 0} \frac{\sqrt{1+2x} – \sqrt{1-2x}}{x}\)
14. \(\displaystyle\lim_{x \to 1} \frac{x^4 – 1}{x^3 – 1}\)
15. \(\displaystyle\lim_{x \to 2} \frac{\sqrt{x+7} – 3}{\sqrt{x+2} – 2}\)
2. Bentuk Tak Tentu \(\dfrac{\infty}{\infty}\)
Materi
Bentuk \(\frac{\infty}{\infty}\) muncul pada limit \(x \to \infty\) untuk fungsi rasional (pecahan polinomial). Strategi:
- Bagi dengan pangkat tertinggi dari penyebut pada pembilang dan penyebut.
- Gunakan aturan cepat:
- Jika derajat pembilang = derajat penyebut → hasilnya = koefisien tertinggi pembilang / koefisien tertinggi penyebut
- Jika derajat pembilang < derajat penyebut → hasilnya = 0
- Jika derajat pembilang > derajat penyebut → hasilnya = ∞ atau −∞
Contoh Soal
Mudah1. Hitung \(\displaystyle\lim_{x \to \infty} \frac{3x + 1}{5x – 2}\)
▶ Lihat PembahasanDerajat sama (1=1). Hasil = \(\frac{3}{5}\)
2. Hitung \(\displaystyle\lim_{x \to \infty} \frac{2x^2}{4x^2 + 1}\)
▶ Lihat PembahasanDerajat sama (2=2). Hasil = \(\frac{2}{4} = \frac{1}{2}\)
3. Hitung \(\displaystyle\lim_{x \to \infty} \frac{x + 5}{x^2 + 1}\)
▶ Lihat PembahasanDerajat pembilang (1) < derajat penyebut (2). Hasil = 0
4. Hitung \(\displaystyle\lim_{x \to \infty} \frac{7x^3}{2x^3 – x}\)
▶ Lihat PembahasanDerajat sama (3=3). Hasil = \(\frac{7}{2}\)
5. Hitung \(\displaystyle\lim_{x \to \infty} \frac{4x}{6x + 3}\)
▶ Lihat PembahasanDerajat sama. Hasil = \(\frac{4}{6} = \frac{2}{3}\)
6. Hitung \(\displaystyle\lim_{x \to \infty} \frac{3x^2 + 2x – 1}{5x^2 – 4x + 7}\)
▶ Lihat PembahasanBagi semua suku dengan \(x^2\): \(\frac{3 + \frac{2}{x} – \frac{1}{x^2}}{5 – \frac{4}{x} + \frac{7}{x^2}}\)
Saat \(x \to \infty\): \(\frac{3+0-0}{5-0+0} = \frac{3}{5}\)
7. Hitung \(\displaystyle\lim_{x \to \infty} \frac{2x^3 – x}{4x^2 + 3}\)
▶ Lihat PembahasanDerajat pembilang (3) > derajat penyebut (2). Hasil = \(\infty\)
Lebih detail: bagi dengan \(x^2\): \(\frac{2x – \frac{1}{x}}{4 + \frac{3}{x^2}} \to \frac{\infty}{4} = \infty\)
8. Hitung \(\displaystyle\lim_{x \to \infty} \frac{x^2 – 3x}{2x^3 + x}\)
▶ Lihat PembahasanDerajat pembilang (2) < penyebut (3). Hasil = 0
9. Hitung \(\displaystyle\lim_{x \to \infty} \frac{(2x+1)(3x-2)}{(x+4)(6x+5)}\)
▶ Lihat PembahasanPembilang: \(6x^2 – x – 2\), Penyebut: \(6x^2 + 29x + 20\)
Derajat sama (2=2). Hasil = \(\frac{6}{6} = 1\)
10. Hitung \(\displaystyle\lim_{x \to \infty} \frac{-5x^2 + x}{3x^2 + 2}\)
▶ Lihat PembahasanDerajat sama. Hasil = \(\frac{-5}{3}\)
11. Hitung \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{3x – 2}\)
▶ Lihat Pembahasan\(\sqrt{4x^2+1} = x\sqrt{4+\frac{1}{x^2}}\) (untuk \(x>0\))
\(\frac{x\sqrt{4+\frac{1}{x^2}}}{3x-2} = \frac{\sqrt{4+\frac{1}{x^2}}}{3-\frac{2}{x}} \to \frac{\sqrt{4}}{3} = \frac{2}{3}\)
12. Hitung \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{9x^2 + x} – 3x}{x + 1}\)
▶ Lihat Pembahasan\(\sqrt{9x^2+x} = x\sqrt{9+\frac{1}{x}} \approx 3x \cdot \sqrt{1+\frac{1}{9x}}\)
Kalikan sekawan: \(\frac{9x^2+x – 9x^2}{(x+1)(\sqrt{9x^2+x}+3x)} = \frac{x}{(x+1)(\sqrt{9x^2+x}+3x)}\)
Bagi dengan \(x\): \(\frac{1}{(1+\frac{1}{x})(\sqrt{9+\frac{1}{x}}+3)} \to \frac{1}{1 \cdot (3+3)} = \frac{1}{6}\)
13. Hitung \(\displaystyle\lim_{x \to \infty} \frac{2x^2 + \sqrt{x^4+1}}{3x^2 – 1}\)
▶ Lihat Pembahasan\(\sqrt{x^4+1} = x^2\sqrt{1+\frac{1}{x^4}} \to x^2\)
\(\frac{2x^2 + x^2}{3x^2} = \frac{3x^2}{3x^2} = 1\)
14. Hitung \(\displaystyle\lim_{x \to \infty} \frac{x^2 + 2^x}{3 \cdot 2^x + x}\)
▶ Lihat PembahasanBagi dengan \(2^x\): \(\frac{\frac{x^2}{2^x} + 1}{3 + \frac{x}{2^x}}\)
Karena eksponensial tumbuh lebih cepat: \(\frac{x^2}{2^x}\to 0\) dan \(\frac{x}{2^x}\to 0\)
Hasil = \(\frac{0+1}{3+0} = \frac{1}{3}\)
15. Hitung \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{x^2+4x} – \sqrt{x^2-2x}}{x}\)
▶ Lihat PembahasanKalikan sekawan pembilang: \(\frac{(x^2+4x)-(x^2-2x)}{x(\sqrt{x^2+4x}+\sqrt{x^2-2x})} = \frac{6x}{x(\sqrt{x^2+4x}+\sqrt{x^2-2x})}\)
\(= \frac{6}{\sqrt{x^2+4x}+\sqrt{x^2-2x}}\). Bagi dalam akar dengan \(x^2\):
\(= \frac{6}{x(\sqrt{1+\frac{4}{x}}+\sqrt{1-\frac{2}{x}})} \cdot x = \frac{6}{\sqrt{1+\frac{4}{x}}+\sqrt{1-\frac{2}{x}}} \to \frac{6}{1+1} = 3\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x \to \infty} \frac{5x – 1}{2x + 3}\)
2. \(\displaystyle\lim_{x \to \infty} \frac{x^2}{3x^2 + 7}\)
3. \(\displaystyle\lim_{x \to \infty} \frac{2x}{x^2 + 1}\)
4. \(\displaystyle\lim_{x \to \infty} \frac{8x^3}{4x^3 – x}\)
5. \(\displaystyle\lim_{x \to \infty} \frac{x + 10}{3x – 7}\)
6. \(\displaystyle\lim_{x \to \infty} \frac{4x^2 – 3x + 2}{2x^2 + x – 5}\)
7. \(\displaystyle\lim_{x \to \infty} \frac{x^3 + 2x}{5x^2 – 1}\)
8. \(\displaystyle\lim_{x \to \infty} \frac{(x+1)(2x-3)}{(3x+2)(x-1)}\)
9. \(\displaystyle\lim_{x \to \infty} \frac{x^2 – 100}{x^2 + 100}\)
10. \(\displaystyle\lim_{x \to \infty} \frac{6x^3 + x}{2x^3 + 3x^2}\)
11. \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{16x^2+3}}{2x – 1}\)
12. \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{x^2+x} – x}{1}\) (sederhanakan dulu)
13. \(\displaystyle\lim_{x \to \infty} \frac{x + \sqrt{x^2+3x}}{2x+1}\)
14. \(\displaystyle\lim_{x \to \infty} \frac{3\sqrt{x^2+1} + x}{5x – 2}\)
15. \(\displaystyle\lim_{x \to \infty} \frac{\sqrt{4x^4 + x^2}}{x^2 + 2x}\)
3. Bentuk Tak Tentu \(0 \cdot \infty\)
Materi
Bentuk \(0 \cdot \infty\) muncul ketika satu faktor mendekati 0 dan faktor lain mendekati ∞. Strategi:
- Ubah ke bentuk \(\frac{0}{0}\) atau \(\frac{\infty}{\infty}\) dengan memindahkan salah satu faktor ke penyebut.
- Contoh: \(f(x) \cdot g(x)\) bisa ditulis \(\frac{f(x)}{1/g(x)}\) atau \(\frac{g(x)}{1/f(x)}\)
Contoh Soal
Mudah1. Hitung \(\displaystyle\lim_{x \to \infty} \frac{1}{x} \cdot x^2\)
▶ Lihat Pembahasan\(\frac{1}{x} \cdot x^2 = x\). Maka \(\displaystyle\lim_{x\to\infty} x = \infty\)
2. Hitung \(\displaystyle\lim_{x \to 0^+} x \cdot \frac{1}{x}\)
▶ Lihat Pembahasan\(x \cdot \frac{1}{x} = 1\). Maka limitnya = 1
3. Hitung \(\displaystyle\lim_{x \to \infty} \frac{3}{x} \cdot (2x+1)\)
▶ Lihat Pembahasan\(\frac{3(2x+1)}{x} = \frac{6x+3}{x} = 6 + \frac{3}{x} \to 6\)
4. Hitung \(\displaystyle\lim_{x \to \infty} \frac{2}{x^2} \cdot x^3\)
▶ Lihat Pembahasan\(\frac{2x^3}{x^2} = 2x \to \infty\)
5. Hitung \(\displaystyle\lim_{x \to \infty} \frac{5}{x} \cdot x\)
▶ Lihat Pembahasan\(\frac{5x}{x} = 5\). Limitnya = 5
6. Hitung \(\displaystyle\lim_{x \to 0^+} x \cdot \ln\frac{1}{x}\)
▶ Lihat Pembahasan\(x \ln\frac{1}{x} = -x\ln x\). Tulis sebagai \(\frac{-\ln x}{1/x}\) (bentuk \(\frac{\infty}{\infty}\))
L’Hôpital: \(\frac{-1/x}{-1/x^2} = \frac{-1/x \cdot x^2}{-1} = x \to 0\)
Hasil = 0
7. Hitung \(\displaystyle\lim_{x \to \infty} \frac{1}{\sqrt{x}} \cdot (x+3)\)
▶ Lihat Pembahasan\(\frac{x+3}{\sqrt{x}} = \frac{x}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x} + \frac{3}{\sqrt{x}} \to \infty\)
8. Hitung \(\displaystyle\lim_{x \to \infty} (x-2)\cdot\frac{4}{x+1}\)
▶ Lihat Pembahasan\(\frac{4(x-2)}{x+1} = \frac{4x-8}{x+1}\). Bagi \(x\): \(\frac{4-\frac{8}{x}}{1+\frac{1}{x}} \to \frac{4}{1} = 4\)
9. Hitung \(\displaystyle\lim_{x \to \infty} \frac{2}{x-1} \cdot (x^2-1)\)
▶ Lihat Pembahasan\(\frac{2(x^2-1)}{x-1} = \frac{2(x-1)(x+1)}{x-1} = 2(x+1) \to \infty\)
10. Hitung \(\displaystyle\lim_{x \to \infty} \frac{x+3}{x^2} \cdot (2x-1)\)
▶ Lihat Pembahasan\(\frac{(x+3)(2x-1)}{x^2} = \frac{2x^2+5x-3}{x^2} = 2 + \frac{5}{x} – \frac{3}{x^2} \to 2\)
11. Hitung \(\displaystyle\lim_{x \to 0^+} \sqrt{x} \cdot \ln x\)
▶ Lihat PembahasanTulis \(\frac{\ln x}{1/\sqrt{x}} = \frac{\ln x}{x^{-1/2}}\) (bentuk \(\frac{-\infty}{\infty}\))
L’Hôpital: \(\frac{1/x}{-\frac{1}{2}x^{-3/2}} = \frac{1/x}{-\frac{1}{2x^{3/2}}} = \frac{x^{3/2}}{x} \cdot (-2) = -2\sqrt{x} \to 0\)
12. Hitung \(\displaystyle\lim_{x \to \infty} x\left(\sqrt{x^2+4} – x\right)\)
▶ Lihat PembahasanKalikan sekawan: \(x \cdot \frac{(x^2+4)-x^2}{\sqrt{x^2+4}+x} = \frac{4x}{\sqrt{x^2+4}+x}\)
Bagi \(x\): \(\frac{4}{\sqrt{1+\frac{4}{x^2}}+1} \to \frac{4}{1+1} = 2\)
13. Hitung \(\displaystyle\lim_{x \to \infty} x^2\left(\frac{1}{\sqrt{x^2+1}} – \frac{1}{\sqrt{x^2+2}}\right)\)
▶ Lihat Pembahasan\(= x^2 \cdot \frac{\sqrt{x^2+2}-\sqrt{x^2+1}}{\sqrt{x^2+1}\cdot\sqrt{x^2+2}}\)
Kalikan sekawan: \(= x^2 \cdot \frac{(x^2+2)-(x^2+1)}{\sqrt{x^2+1}\sqrt{x^2+2}(\sqrt{x^2+2}+\sqrt{x^2+1})}\)
\(= \frac{x^2}{\sqrt{x^2+1}\sqrt{x^2+2}(\sqrt{x^2+2}+\sqrt{x^2+1})}\)
Bagi \(x^2\) dan \(x\): \(\to \frac{1}{1\cdot1\cdot(1+1)} = \frac{1}{2}\) — wait, more carefully:
Denom \(\approx x \cdot x \cdot 2x = 2x^3\). So \(\frac{x^2}{2x^3} = \frac{1}{2x}\to 0\)
14. Hitung \(\displaystyle\lim_{x \to \infty} (2x+3)\left(\frac{1}{\sqrt{4x^2+x}} – \frac{1}{2x}\right)\)
▶ Lihat Pembahasan\(= (2x+3)\cdot\frac{2x – \sqrt{4x^2+x}}{2x\sqrt{4x^2+x}}\)
Sekawan: \(2x-\sqrt{4x^2+x} = \frac{4x^2-(4x^2+x)}{2x+\sqrt{4x^2+x}} = \frac{-x}{2x+\sqrt{4x^2+x}}\)
\(= \frac{(2x+3)(-x)}{2x\sqrt{4x^2+x}(2x+\sqrt{4x^2+x})}\)
Orde: pembilang ~\(2x^2\), penyebut ~ \(2x \cdot 2x \cdot 4x = 16x^3\). Hasilnya \(\to 0\)
Lebih detail: bagi semua dengan \(x^2\): \(\frac{-(2+3/x)}{2\sqrt{4+1/x}(2+\sqrt{4+1/x})} \to \frac{-2}{2\cdot2\cdot(2+2)} = \frac{-2}{16} = -\frac{1}{8}\)
15. Hitung \(\displaystyle\lim_{x \to \infty} x\left(\sqrt{x^2+x+1} – \sqrt{x^2-x+1}\right)\)
▶ Lihat PembahasanSekawan: \(\frac{(x^2+x+1)-(x^2-x+1)}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}} = \frac{2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}\)
Maka: \(\frac{x \cdot 2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}} = \frac{2x^2}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}\)
Bagi \(x\): \(\frac{2x}{\sqrt{1+1/x+1/x^2}+\sqrt{1-1/x+1/x^2}} \to \frac{2x}{1+1} = x \to \infty\)
Hmm, cek ulang. \(\frac{2x^2}{x(\sqrt{1+1/x+1/x^2}+\sqrt{1-1/x+1/x^2})} = \frac{2x}{2} = x\to\infty\)
Jadi hasilnya = \(\infty\). Namun soalnya bentuk \(0\cdot\infty\)? Karena di luar \(x\to\infty\), selisih akar \(\to 1\), jadi ini sebenarnya \(\infty\cdot 1 = \infty\). Mari koreksi: \(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\to 1\), maka \(x\cdot 1 \to \infty\). Jawab: \(\infty\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x \to \infty} \frac{7}{x}\cdot(x+2)\)
2. \(\displaystyle\lim_{x \to \infty} \frac{4}{x^2}\cdot x^3\)
3. \(\displaystyle\lim_{x \to \infty} \frac{1}{x+1}\cdot(3x)\)
4. \(\displaystyle\lim_{x \to \infty} \frac{6}{2x-1}\cdot x\)
5. \(\displaystyle\lim_{x \to \infty} \frac{x-1}{x^2}\cdot(2x)\)
6. \(\displaystyle\lim_{x \to \infty} (x+5)\cdot\frac{3}{x-1}\)
7. \(\displaystyle\lim_{x \to \infty} \frac{x^2-4}{x^3}\cdot(x+2)\)
8. \(\displaystyle\lim_{x \to 0^+} x^2 \cdot \frac{1}{x}\)
9. \(\displaystyle\lim_{x \to \infty} \frac{2x+1}{x^2+1}\cdot(x-3)\)
10. \(\displaystyle\lim_{x \to \infty} (3x-2)\cdot\frac{x}{x^2+5}\)
11. \(\displaystyle\lim_{x \to 0^+} x^2\ln x\)
12. \(\displaystyle\lim_{x \to \infty} x(\sqrt{x^2+9}-x)\)
13. \(\displaystyle\lim_{x \to \infty} x(\sqrt{4x^2+1}-2x)\)
14. \(\displaystyle\lim_{x \to \infty} (x+1)\left(\frac{1}{\sqrt{x^2+3}}-\frac{1}{x}\right)\)
15. \(\displaystyle\lim_{x \to \infty} x^2\left(\sqrt{1+\frac{2}{x}}-\sqrt{1-\frac{2}{x}}\right)\)
4. Bentuk Tak Tentu \(\infty – \infty\)
Materi
Bentuk \(\infty – \infty\) terjadi ketika limit menghasilkan selisih dua besaran yang keduanya menuju tak hingga. Strategi:
- Kalikan sekawan — jika melibatkan akar, kalikan dengan \(\frac{\text{sekawan}}{\text{sekawan}}\).
- Samakan penyebut — jika melibatkan pecahan, samakan penyebut lalu sederhanakan.
- Gunakan rumus cepat: \(\displaystyle\lim_{x\to\infty}(\sqrt{ax^2+bx+c}-\sqrt{ax^2+dx+e}) = \frac{b-d}{2\sqrt{a}}\)
Contoh Soal
Mudah1. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+4x} – x)\)
▶ Lihat PembahasanTulis \(\sqrt{x^2+4x} – \sqrt{x^2}\). Rumus cepat: \(a=1, b=4, d=0\)
Hasil = \(\frac{4-0}{2\sqrt{1}} = \frac{4}{2} = 2\)
Atau: sekawan \(\frac{x^2+4x-x^2}{\sqrt{x^2+4x}+x} = \frac{4x}{\sqrt{x^2+4x}+x}\). Bagi \(x\): \(\frac{4}{\sqrt{1+4/x}+1}\to\frac{4}{2}=2\)
2. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+6x} – x)\)
▶ Lihat PembahasanRumus cepat: \(\frac{6-0}{2\cdot 1} = 3\)
3. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+2x+1} – x)\)
▶ Lihat Pembahasan\(\sqrt{x^2+2x+1} = \sqrt{(x+1)^2} = |x+1| = x+1\) (untuk \(x>0\))
\((x+1) – x = 1\). Jawab: 1
4. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+10x} – x)\)
▶ Lihat PembahasanRumus cepat: \(\frac{10}{2} = 5\)
5. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{4x^2+x} – 2x)\)
▶ Lihat PembahasanSekawan: \(\frac{4x^2+x-4x^2}{\sqrt{4x^2+x}+2x} = \frac{x}{\sqrt{4x^2+x}+2x}\)
Bagi \(x\): \(\frac{1}{\sqrt{4+1/x}+2} \to \frac{1}{2+2} = \frac{1}{4}\)
6. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+3x} – \sqrt{x^2+x})\)
▶ Lihat PembahasanRumus cepat: \(a=1, b=3, d=1\). Hasil = \(\frac{3-1}{2\sqrt{1}} = 1\)
7. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{4x^2+8x+1} – 2x)\)
▶ Lihat PembahasanSekawan: \(\frac{4x^2+8x+1-4x^2}{\sqrt{4x^2+8x+1}+2x} = \frac{8x+1}{\sqrt{4x^2+8x+1}+2x}\)
Bagi \(x\): \(\frac{8+1/x}{\sqrt{4+8/x+1/x^2}+2} \to \frac{8}{2+2} = 2\)
8. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{9x^2+12x} – 3x)\)
▶ Lihat PembahasanSekawan: \(\frac{9x^2+12x-9x^2}{\sqrt{9x^2+12x}+3x} = \frac{12x}{\sqrt{9x^2+12x}+3x}\)
Bagi \(x\): \(\frac{12}{\sqrt{9+12/x}+3}\to\frac{12}{3+3} = 2\)
9. Hitung \(\displaystyle\lim_{x \to \infty} \left(\frac{x^2}{x-1} – \frac{x^2}{x+1}\right)\)
▶ Lihat PembahasanSamakan penyebut: \(\frac{x^2(x+1)-x^2(x-1)}{(x-1)(x+1)} = \frac{x^3+x^2-x^3+x^2}{x^2-1} = \frac{2x^2}{x^2-1}\)
Bagi \(x^2\): \(\frac{2}{1-1/x^2} \to 2\)
10. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+5x+2} – \sqrt{x^2+3x+1})\)
▶ Lihat PembahasanRumus cepat: \(\frac{5-3}{2\sqrt{1}} = 1\)
11. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{4x^2+3x} – \sqrt{4x^2-x+2})\)
▶ Lihat PembahasanRumus cepat: \(a=4, b=3, d=-1\). Hasil = \(\frac{3-(-1)}{2\sqrt{4}} = \frac{4}{4} = 1\)
12. Hitung \(\displaystyle\lim_{x \to \infty} \left(\frac{x^3}{x^2-1} – \frac{x^3}{x^2+1}\right)\)
▶ Lihat Pembahasan\(\frac{x^3(x^2+1)-x^3(x^2-1)}{(x^2-1)(x^2+1)} = \frac{2x^3}{x^4-1}\)
Bagi \(x^4\): \(\frac{2/x}{1-1/x^4} \to 0\)
13. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt[3]{x^3+3x^2} – x)\)
▶ Lihat PembahasanMisalkan \(y = \sqrt[3]{x^3+3x^2} – x\). Maka \(y+x = \sqrt[3]{x^3+3x^2}\), jadi \((y+x)^3 = x^3+3x^2\).
Gunakan identitas \(a-b = \frac{a^3-b^3}{a^2+ab+b^2}\) dengan \(a=\sqrt[3]{x^3+3x^2}, b=x\):
\(\frac{x^3+3x^2-x^3}{(\sqrt[3]{x^3+3x^2})^2+x\sqrt[3]{x^3+3x^2}+x^2} = \frac{3x^2}{3x^2(\text{suku dominan})} = 1\)
Lebih detail: penyebut ~\(x^2+x^2+x^2 = 3x^2\). Jawab: \(\frac{3x^2}{3x^2} = 1\)
14. Hitung \(\displaystyle\lim_{x \to \infty} \left(\frac{2x^2+x}{2x-1} – x\right)\)
▶ Lihat Pembahasan\(\frac{2x^2+x}{2x-1} – x = \frac{2x^2+x-x(2x-1)}{2x-1} = \frac{2x^2+x-2x^2+x}{2x-1} = \frac{2x}{2x-1}\)
Bagi \(x\): \(\frac{2}{2-1/x}\to 1\)
15. Hitung \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+ax} – \sqrt{x^2+bx})\) dalam \(a\) dan \(b\)
▶ Lihat PembahasanRumus cepat langsung: \(\frac{a-b}{2\sqrt{1}} = \frac{a-b}{2}\)
Latihan Soal
Mudah1. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+8x} – x)\)
2. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+12x+5} – x)\)
3. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+2x} – x)\)
4. \(\displaystyle\lim_{x \to \infty} (\sqrt{9x^2+6x} – 3x)\)
5. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+20x} – x)\)
6. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+7x} – \sqrt{x^2+3x})\)
7. \(\displaystyle\lim_{x \to \infty} (\sqrt{4x^2+4x+1} – 2x)\)
8. \(\displaystyle\lim_{x \to \infty} \left(\frac{x^2}{x-2} – \frac{x^2}{x+2}\right)\)
9. \(\displaystyle\lim_{x \to \infty} (\sqrt{x^2+4x-1} – \sqrt{x^2-2x+3})\)
10. \(\displaystyle\lim_{x \to \infty} (\sqrt{16x^2+x} – 4x)\)
11. \(\displaystyle\lim_{x \to \infty} (\sqrt{9x^2+5x} – \sqrt{9x^2-x})\)
12. \(\displaystyle\lim_{x \to \infty} \left(\frac{3x^2+x}{3x-2} – x\right)\)
13. \(\displaystyle\lim_{x \to \infty} (\sqrt[3]{x^3+6x^2} – x)\)
14. \(\displaystyle\lim_{x \to \infty} \left(\frac{x^3+1}{x^2+1} – x\right)\)
15. \(\displaystyle\lim_{x \to \infty} (\sqrt{25x^2+3x+1} – \sqrt{25x^2-2x})\)
Limit Fungsi yang Mengarah ke Konsep Turunan
Limit Fungsi yang Mengarah ke Konsep Turunan
Matematika Kelas XI — Materi, Contoh Soal & Latihan
1. Pengertian Limit Fungsi
Limit fungsi menyatakan nilai yang didekati oleh suatu fungsi ketika variabelnya mendekati suatu nilai tertentu. Secara formal:
Artinya: ketika \(x\) mendekati \(a\), nilai \(f(x)\) mendekati \(L\).
2. Limit dan Konsep Turunan (Derivatif)
Turunan fungsi \(f(x)\) di titik \(x = a\) didefinisikan sebagai limit dari laju perubahan rata-rata:
Atau bentuk ekuivalen:
Secara geometris, turunan adalah gradien garis singgung kurva \(y = f(x)\) di titik \((a, f(a))\).
3. Ilustrasi Geometris
Perhatikan ilustrasi berikut yang menunjukkan garis potong mendekati garis singgung:
Ketika \(h \to 0\), garis potong (kuning putus-putus) mendekati garis singgung (hijau).
4. Tabel Pendekatan Nilai Limit
Contoh: \(f(x) = x^2\), cari \(f'(2)\) menggunakan tabel pendekatan \(h \to 0\):
| \(h\) | \(f(2+h)\) | \(\frac{f(2+h)-f(2)}{h}\) |
|---|---|---|
| 1 | 9 | 5 |
| 0.1 | 4.41 | 4.1 |
| 0.01 | 4.0401 | 4.01 |
| 0.001 | 4.004001 | 4.001 |
| → 0 | → 4 | → 4 |
Kesimpulan: \(f'(2) = \lim_{h \to 0} \frac{(2+h)^2 – 4}{h} = 4\)
5. Rumus-Rumus Turunan Dasar
Contoh Soal & Pembahasan
MUDAHContoh Soal Tingkat Mudah
Contoh 1
Tentukan \(f'(x)\) dari \(f(x) = 3x + 5\) menggunakan definisi limit turunan.
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}\]
\[= \lim_{h \to 0} \frac{[3(x+h)+5] – [3x+5]}{h}\]
\[= \lim_{h \to 0} \frac{3x + 3h + 5 – 3x – 5}{h}\]
\[= \lim_{h \to 0} \frac{3h}{h} = \lim_{h \to 0} 3 = 3\]
Jawaban: \(f'(x) = 3\)
Contoh 2
Tentukan \(f'(x)\) dari \(f(x) = x^2\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0} \frac{(x+h)^2 – x^2}{h}\]
\[= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 – x^2}{h}\]
\[= \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x\]
Jawaban: \(f'(x) = 2x\)
Contoh 3
Tentukan \(f'(1)\) jika \(f(x) = 4x – 1\).
\[f'(1) = \lim_{h \to 0} \frac{f(1+h) – f(1)}{h}\]
\[= \lim_{h \to 0} \frac{[4(1+h)-1] – [4(1)-1]}{h}\]
\[= \lim_{h \to 0} \frac{4+4h-1-3}{h} = \lim_{h \to 0} \frac{4h}{h} = 4\]
Jawaban: \(f'(1) = 4\)
Contoh 4
Tentukan \(f'(x)\) dari \(f(x) = 5\) (fungsi konstan).
\[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{5-5}{h} = \lim_{h \to 0} \frac{0}{h} = 0\]
Jawaban: \(f'(x) = 0\)
Contoh 5
Tentukan \(f'(x)\) dari \(f(x) = 2x^2 + 1\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0} \frac{[2(x+h)^2+1]-[2x^2+1]}{h}\]
\[= \lim_{h \to 0} \frac{2x^2+4xh+2h^2+1-2x^2-1}{h}\]
\[= \lim_{h \to 0} \frac{4xh+2h^2}{h} = \lim_{h \to 0}(4x+2h) = 4x\]
Jawaban: \(f'(x) = 4x\)
SEDANGContoh Soal Tingkat Sedang
Contoh 6
Tentukan \(f'(x)\) dari \(f(x) = x^3\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0} \frac{(x+h)^3 – x^3}{h}\]
Ekspansi: \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\)
\[= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2\]
Jawaban: \(f'(x) = 3x^2\)
Contoh 7
Tentukan \(f'(x)\) dari \(f(x) = \frac{1}{x}\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} – \frac{1}{x}}{h}\]
\[= \lim_{h \to 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)}\]
\[= \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x^2}\]
Jawaban: \(f'(x) = -\frac{1}{x^2}\)
Contoh 8
Tentukan \(f'(x)\) dari \(f(x) = \sqrt{x}\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} – \sqrt{x}}{h}\]
Kalikan sekawan:
\[= \lim_{h \to 0} \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\]
\[= \lim_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\]
Jawaban: \(f'(x) = \frac{1}{2\sqrt{x}}\)
Contoh 9
Tentukan gradien garis singgung kurva \(y = x^2 – 3x + 2\) di titik \(x = 2\).
\[f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}\]
\(f(2) = 4-6+2 = 0\)
\(f(2+h) = (2+h)^2 – 3(2+h)+2 = 4+4h+h^2-6-3h+2 = h^2+h\)
\[f'(2) = \lim_{h \to 0} \frac{h^2+h-0}{h} = \lim_{h \to 0}(h+1) = 1\]
Jawaban: Gradien = 1
Contoh 10
Tentukan \(f'(x)\) dari \(f(x) = x^2 + 4x\) menggunakan definisi limit, lalu hitung \(f'(3)\).
\[f'(x) = \lim_{h \to 0}\frac{[(x+h)^2+4(x+h)]-[x^2+4x]}{h}\]
\[= \lim_{h \to 0}\frac{2xh+h^2+4h}{h} = \lim_{h \to 0}(2x+h+4) = 2x+4\]
\(f'(3) = 2(3)+4 = 10\)
Jawaban: \(f'(x) = 2x+4\), \(f'(3) = 10\)
SULITContoh Soal Tingkat Sulit
Contoh 11
Tentukan \(f'(x)\) dari \(f(x) = \frac{x}{x+1}\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0}\frac{\frac{x+h}{x+h+1} – \frac{x}{x+1}}{h}\]
Samakan penyebut pembilang:
\[= \lim_{h \to 0}\frac{(x+h)(x+1) – x(x+h+1)}{h(x+h+1)(x+1)}\]
\[= \lim_{h \to 0}\frac{x^2+x+hx+h – x^2-hx-x}{h(x+h+1)(x+1)}\]
\[= \lim_{h \to 0}\frac{h}{h(x+h+1)(x+1)} = \frac{1}{(x+1)^2}\]
Jawaban: \(f'(x) = \frac{1}{(x+1)^2}\)
Contoh 12
Tentukan \(f'(x)\) dari \(f(x) = \sqrt{2x+1}\) menggunakan definisi limit.
\[f'(x) = \lim_{h \to 0}\frac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}\]
Kalikan sekawan:
\[= \lim_{h \to 0}\frac{[2(x+h)+1]-[2x+1]}{h[\sqrt{2x+2h+1}+\sqrt{2x+1}]}\]
\[= \lim_{h \to 0}\frac{2h}{h[\sqrt{2x+2h+1}+\sqrt{2x+1}]} = \frac{2}{2\sqrt{2x+1}} = \frac{1}{\sqrt{2x+1}}\]
Jawaban: \(f'(x) = \frac{1}{\sqrt{2x+1}}\)
Contoh 13
Tentukan \(f'(x)\) dari \(f(x) = x^2 \cdot \sqrt{x}\) menggunakan definisi limit di \(x = 4\).
Tulis ulang: \(f(x) = x^{5/2}\)
\[f'(4) = \lim_{x \to 4}\frac{x^{5/2}-4^{5/2}}{x-4} = \lim_{x \to 4}\frac{x^{5/2}-32}{x-4}\]
Substitusi \(u = \sqrt{x}\), jadi \(x = u^2\), \(x \to 4 \Rightarrow u \to 2\):
\[= \lim_{u \to 2}\frac{u^5-32}{u^2-4} = \lim_{u \to 2}\frac{(u-2)(u^4+2u^3+4u^2+8u+16)}{(u-2)(u+2)}\]
\[= \frac{16+16+16+16+16}{4} = \frac{80}{4} = 20\]
Jawaban: \(f'(4) = 20\)
Contoh 14
Tentukan persamaan garis singgung kurva \(y = \frac{1}{x^2}\) di titik \(x = 1\).
Titik singgung: \((1, 1)\)
\[f'(1) = \lim_{h \to 0}\frac{\frac{1}{(1+h)^2}-1}{h} = \lim_{h \to 0}\frac{1-(1+h)^2}{h(1+h)^2}\]
\[= \lim_{h \to 0}\frac{1-1-2h-h^2}{h(1+h)^2} = \lim_{h \to 0}\frac{-2h-h^2}{h(1+h)^2}\]
\[= \lim_{h \to 0}\frac{-2-h}{(1+h)^2} = \frac{-2}{1} = -2\]
Persamaan garis singgung: \(y – 1 = -2(x-1)\) → \(y = -2x + 3\)
Jawaban: \(y = -2x + 3\)
Contoh 15
Jika \(f(x) = x^3 – 2x^2 + x\), tentukan titik-titik di mana garis singgung horizontal (gradien = 0).
Cari \(f'(x)\) menggunakan limit:
\[f'(x) = \lim_{h \to 0}\frac{[(x+h)^3-2(x+h)^2+(x+h)]-[x^3-2x^2+x]}{h}\]
Setelah ekspansi dan penyederhanaan:
\[f'(x) = 3x^2 – 4x + 1\]
Syarat gradien = 0: \(3x^2 – 4x + 1 = 0\)
\((3x-1)(x-1) = 0\) → \(x = \frac{1}{3}\) atau \(x = 1\)
\(f(\frac{1}{3}) = \frac{1}{27}-\frac{2}{9}+\frac{1}{3} = \frac{4}{27}\)
\(f(1) = 1-2+1 = 0\)
Jawaban: Titik \((\frac{1}{3}, \frac{4}{27})\) dan \((1, 0)\)
Latihan Soal
Kerjakan soal-soal berikut tanpa melihat pembahasan contoh soal di atas.
MUDAHLatihan Tingkat Mudah
1. Tentukan \(f'(x)\) dari \(f(x) = 7x – 2\) menggunakan definisi limit turunan.
2. Tentukan \(f'(x)\) dari \(f(x) = -3x + 10\) menggunakan definisi limit turunan.
3. Tentukan \(f'(2)\) jika \(f(x) = x^2 + 1\).
4. Tentukan \(f'(x)\) dari \(f(x) = 5x^2\) menggunakan definisi limit.
5. Tentukan \(f'(0)\) jika \(f(x) = x^2 – 4x + 3\).
SEDANGLatihan Tingkat Sedang
6. Tentukan \(f'(x)\) dari \(f(x) = x^3 + 2x\) menggunakan definisi limit.
7. Tentukan \(f'(x)\) dari \(f(x) = \frac{2}{x}\) menggunakan definisi limit.
8. Tentukan gradien garis singgung kurva \(y = x^2 + 2x – 1\) di titik \(x = 3\).
9. Tentukan \(f'(x)\) dari \(f(x) = \sqrt{3x}\) menggunakan definisi limit.
10. Tentukan \(f'(x)\) dari \(f(x) = x^2 – \frac{1}{x}\) menggunakan definisi limit.
SULITLatihan Tingkat Sulit
11. Tentukan \(f'(x)\) dari \(f(x) = \frac{2x+1}{x-1}\) menggunakan definisi limit.
12. Tentukan \(f'(x)\) dari \(f(x) = \sqrt{x^2+1}\) menggunakan definisi limit.
13. Tentukan persamaan garis singgung kurva \(y = x^3 – x\) di titik \((1, 0)\).
14. Tentukan \(f'(x)\) dari \(f(x) = \frac{1}{\sqrt{x+2}}\) menggunakan definisi limit.
15. Jika \(f(x) = x^4 – 4x^2\), tentukan semua titik di mana garis singgung kurva memiliki gradien = 0.