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. \(\displaystyle\lim_{x\to\infty}x(\sqrt{x^2+4}-x)\)