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. \(\displaystyle\lim_{x \to \infty} (\sqrt{25x^2+3x+1} – \sqrt{25x^2-2x})\)