以下の6つの極限値を求める問題です。 (1) $\lim_{x\to 0} \frac{\sin 4x}{\sin 5x}$ (2) $\lim_{x\to 0} \frac{e^{3x}-e^x}{\sin^{-1} x}$ (3) $\lim_{x\to 0} \frac{\sqrt{1+x^2}-1}{(1+x)^{1/3} - x/3 - 1}$ (4) $\lim_{x\to 0} \frac{\log(\cos x)}{\sin^2 x}$ (5) $\lim_{x\to 0} (\frac{1}{\log(1+x)} - \frac{1}{x})$ (6) $\lim_{x\to 0} \frac{\tan x - \sin^{-1} x}{e^x + \log(1-x) - 1}$

1. 問題の内容

以下の6つの極限値を求める問題です。

(1) $\lim_{x\to 0} \frac{\sin 4x}{\sin 5x}$

(2) $\lim_{x\to 0} \frac{e^{3x}-e^x}{\sin^{-1} x}$

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

(4) $\lim_{x\to 0} \frac{\log(\cos x)}{\sin^2 x}$

(5) $\lim_{x\to 0} (\frac{1}{\log(1+x)} - \frac{1}{x})$

(6) $\lim_{x\to 0} \frac{\tan x - \sin^{-1} x}{e^x + \log(1-x) - 1}$

2. 解き方の手順

(1) $\lim_{x\to 0} \frac{\sin 4x}{\sin 5x} = \lim_{x\to 0} \frac{\sin 4x}{4x} \cdot \frac{5x}{\sin 5x} \cdot \frac{4x}{5x} = 1 \cdot 1 \cdot \frac{4}{5} = \frac{4}{5}$

(2) $\lim_{x\to 0} \frac{e^{3x}-e^x}{\sin^{-1} x} = \lim_{x\to 0} \frac{e^{3x}-1-(e^x-1)}{\sin^{-1} x} = \lim_{x\to 0} \frac{e^{3x}-1}{3x} \cdot \frac{x}{\sin^{-1} x} \cdot 3 - \frac{e^x-1}{x} \cdot \frac{x}{\sin^{-1} x} \cdot 1 = 1 \cdot 1 \cdot 3 - 1 \cdot 1 \cdot 1 = 3 - 1 = 2$

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

$(1+x)^{1/3} = 1 + \frac{1}{3}x + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}x^2 + O(x^3) = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + O(x^3)$

$(1+x)^{1/3} - x/3 - 1 = 1 + \frac{1}{3}x - \frac{1}{9}x^2 - \frac{x}{3} - 1 + O(x^3) = -\frac{1}{9}x^2 + O(x^3)$

よって、$\lim_{x\to 0} \frac{x^2}{((1+x)^{1/3} - x/3 - 1)(\sqrt{1+x^2}+1)} = \lim_{x\to 0} \frac{x^2}{(-\frac{1}{9}x^2)(2)} = -\frac{9}{2}$

(4) $\lim_{x\to 0} \frac{\log(\cos x)}{\sin^2 x} = \lim_{x\to 0} \frac{\log(1+(\cos x - 1))}{\sin^2 x} = \lim_{x\to 0} \frac{\cos x - 1}{\sin^2 x} = \lim_{x\to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2}$

(5) $\lim_{x\to 0} (\frac{1}{\log(1+x)} - \frac{1}{x}) = \lim_{x\to 0} \frac{x - \log(1+x)}{x \log(1+x)} = \lim_{x\to 0} \frac{x - (x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)}{x(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = \lim_{x\to 0} \frac{\frac{x^2}{2} - \frac{x^3}{3} + \dots}{x^2 - \frac{x^3}{2} + \dots} = \frac{1}{2}$

(6) $\lim_{x\to 0} \frac{\tan x - \sin^{-1} x}{e^x + \log(1-x) - 1} = \lim_{x\to 0} \frac{(x + \frac{x^3}{3} + \dots) - (x + \frac{x^3}{6} + \dots)}{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots) + (-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots) - 1} = \lim_{x\to 0} \frac{\frac{x^3}{6} + \dots}{-\frac{x^3}{6} + \dots} = -1$

3. 最終的な答え

(1) $\frac{4}{5}$

(2) $2$

(3) $-\frac{9}{2}$

(4) $-\frac{1}{2}$

(5) $\frac{1}{2}$

(6) $-1$