次の3つの極限値を求める問題です。 (1) $\lim_{x\to 0} \frac{\sin(-1 + \cos(\cos(x) - 1))}{x^4}$ (2) $\lim_{x\to \infty} \frac{\sqrt{x^2 + x + 1} - \sqrt{x^2}}{(\sqrt[3]{x+1} - \sqrt[3]{x})x^{2/3}}$ (3) $\lim_{x\to \infty} (3^x + 5^x)^{\frac{1}{x-1}}$

1. 問題の内容

次の3つの極限値を求める問題です。

(1) $\lim_{x\to 0} \frac{\sin(-1 + \cos(\cos(x) - 1))}{x^4}$

(2) $\lim_{x\to \infty} \frac{\sqrt{x^2 + x + 1} - \sqrt{x^2}}{(\sqrt[3]{x+1} - \sqrt[3]{x})x^{2/3}}$

(3) $\lim_{x\to \infty} (3^x + 5^x)^{\frac{1}{x-1}}$

2. 解き方の手順

(1)

$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$ より、

$\cos(x) - 1 = -\frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$

$\cos(\cos(x) - 1) = 1 - \frac{1}{2}(-\frac{x^2}{2} + \frac{x^4}{24})^2 + O(x^6)$

$\cos(\cos(x) - 1) - 1 = -\frac{1}{2}(\frac{x^4}{4}) + O(x^6) = -\frac{x^4}{8} + O(x^6)$

$-1 + \cos(\cos(x) - 1) = -\frac{x^4}{8} + O(x^6)$

$\sin(x) = x + O(x^3)$ を用いると、

$\sin(-1 + \cos(\cos(x) - 1)) = -\frac{x^4}{8} + O(x^6)$

$\lim_{x\to 0} \frac{\sin(-1 + \cos(\cos(x) - 1))}{x^4} = \lim_{x\to 0} \frac{-\frac{x^4}{8} + O(x^6)}{x^4} = -\frac{1}{8}$

(2)

$\sqrt{x^2 + x + 1} - \sqrt{x^2} = \frac{x^2 + x + 1 - x^2}{\sqrt{x^2 + x + 1} + \sqrt{x^2}} = \frac{x + 1}{\sqrt{x^2 + x + 1} + x} = \frac{x(1 + \frac{1}{x})}{x(\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1)} = \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1}$

$\sqrt[3]{x+1} - \sqrt[3]{x} = \frac{(x+1) - x}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\sqrt[3]{x} + (\sqrt[3]{x})^2} = \frac{1}{(x+1)^{2/3} + (x^2 + x)^{1/3} + x^{2/3}} = \frac{1}{x^{2/3}((1+\frac{1}{x})^{2/3} + (1+\frac{1}{x})^{1/3} + 1)}$

$\frac{\sqrt{x^2 + x + 1} - \sqrt{x^2}}{(\sqrt[3]{x+1} - \sqrt[3]{x})x^{2/3}} = \frac{\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1}}{\frac{1}{x^{2/3}((1+\frac{1}{x})^{2/3} + (1+\frac{1}{x})^{1/3} + 1)}x^{2/3}} = \frac{(1 + \frac{1}{x})((1+\frac{1}{x})^{2/3} + (1+\frac{1}{x})^{1/3} + 1)}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1}$

$\lim_{x\to \infty} \frac{\sqrt{x^2 + x + 1} - \sqrt{x^2}}{(\sqrt[3]{x+1} - \sqrt[3]{x})x^{2/3}} = \frac{(1+0)(1+1+1)}{\sqrt{1+0+0} + 1} = \frac{3}{2}$

(3)

$\lim_{x\to \infty} (3^x + 5^x)^{\frac{1}{x-1}} = \lim_{x\to \infty} (5^x(\frac{3^x}{5^x} + 1))^{\frac{1}{x-1}} = \lim_{x\to \infty} 5^{\frac{x}{x-1}} (\frac{3^x}{5^x} + 1)^{\frac{1}{x-1}} = \lim_{x\to \infty} 5^{\frac{x}{x-1}} ((\frac{3}{5})^x + 1)^{\frac{1}{x-1}}$

$\lim_{x\to \infty} \frac{x}{x-1} = 1$

$\lim_{x\to \infty} (\frac{3}{5})^x = 0$

$\lim_{x\to \infty} (0 + 1)^{\frac{1}{x-1}} = 1^0 = 1$

よって、$\lim_{x\to \infty} (3^x + 5^x)^{\frac{1}{x-1}} = 5^1 \cdot 1 = 5$

3. 最終的な答え

(1) $-\frac{1}{8}$

(2) $\frac{3}{2}$

(3) $5$