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

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

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

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

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

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

$\lim_{x \to 0} \frac{x}{\sin^{-1} x} = 1$

よって、$\lim_{x \to 0} \frac{e^{3x} - e^x}{\sin^{-1} x} = 2 \cdot 1 = 2$

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

$\sqrt{1+x^2} = 1 + \frac{1}{2}x^2 + O(x^4)$

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

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

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

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

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

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

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

$\sin x = x + O(x^3)$

$\sin^2 x = x^2 + O(x^4)$

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

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

$\lim_{x \to 0} \left( \frac{1}{\log(1+x)} - \frac{1}{x} \right) = \lim_{x \to 0} \frac{x - \log(1+x)}{x \log(1+x)}$

$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$

$x - \log(1+x) = \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)$

$x \log(1+x) = x(x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)) = x^2 - \frac{x^3}{2} + O(x^4)$

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

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

$\tan x = x + \frac{x^3}{3} + O(x^5)$

$\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$

$\log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4)$

$\tan x - \sin^{-1} x = \frac{x^3}{3} - \frac{x^3}{6} + O(x^5) = \frac{x^3}{6} + O(x^5)$

$e^x + \log(1-x) - 1 = (1 + x + \frac{x^2}{2} + \frac{x^3}{6}) + (-x - \frac{x^2}{2} - \frac{x^3}{3}) - 1 + O(x^4) = -\frac{x^3}{6} + O(x^4)$

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

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) $-1$