次の極限値を求めます。 (1) $\lim_{x \to 0} \frac{\tan x}{x + \sin x}$ (4) $\lim_{x \to 0} \frac{e^{2x} - 1}{x^2 + x}$ (7) $\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}$ (10) $\lim_{x \to \frac{\pi}{3}} \frac{\tan x - \sqrt{3}}{\sin(x - \frac{\pi}{3})}$

1. 問題の内容

次の極限値を求めます。

(1) $\lim_{x \to 0} \frac{\tan x}{x + \sin x}$

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

(7) $\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}$

(10) $\lim_{x \to \frac{\pi}{3}} \frac{\tan x - \sqrt{3}}{\sin(x - \frac{\pi}{3})}$

2. 解き方の手順

(1) $\lim_{x \to 0} \frac{\tan x}{x + \sin x}$

$\tan x = \frac{\sin x}{\cos x}$ なので、与式は

$\lim_{x \to 0} \frac{\frac{\sin x}{\cos x}}{x + \sin x} = \lim_{x \to 0} \frac{\sin x}{\cos x (x + \sin x)}$

分母分子を $x$ で割ると

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

$\lim_{x \to 0} \frac{\sin x}{x} = 1$、$\lim_{x \to 0} \cos x = 1$ なので、

$\frac{1}{1 \cdot (1 + 1)} = \frac{1}{2}$

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

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

$\lim_{x \to 0} \frac{e^{2x} - 1}{2x} \cdot \frac{2}{x+1}$

$\lim_{x \to 0} \frac{e^{2x} - 1}{2x} = 1$ なので、

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

(7) $\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}$

$\lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \cdot \frac{5x}{2x}$

$\lim_{x \to 0} \frac{\sin 5x}{5x} = 1$, $\lim_{x \to 0} \frac{\sin 2x}{2x} = 1$ なので、

$1 \cdot 1 \cdot \frac{5}{2} = \frac{5}{2}$

(10) $\lim_{x \to \frac{\pi}{3}} \frac{\tan x - \sqrt{3}}{\sin(x - \frac{\pi}{3})}$

$t = x - \frac{\pi}{3}$ とおくと、$x \to \frac{\pi}{3}$ のとき $t \to 0$ なので、

$\lim_{t \to 0} \frac{\tan (t + \frac{\pi}{3}) - \sqrt{3}}{\sin t}$

$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$ より、

$\tan(t + \frac{\pi}{3}) = \frac{\tan t + \tan \frac{\pi}{3}}{1 - \tan t \tan \frac{\pi}{3}} = \frac{\tan t + \sqrt{3}}{1 - \sqrt{3} \tan t}$

よって、$\lim_{t \to 0} \frac{\frac{\tan t + \sqrt{3}}{1 - \sqrt{3} \tan t} - \sqrt{3}}{\sin t} = \lim_{t \to 0} \frac{\tan t + \sqrt{3} - \sqrt{3}(1 - \sqrt{3} \tan t)}{\sin t (1 - \sqrt{3} \tan t)} = \lim_{t \to 0} \frac{\tan t + 3 \tan t}{\sin t (1 - \sqrt{3} \tan t)} = \lim_{t \to 0} \frac{4 \tan t}{\sin t (1 - \sqrt{3} \tan t)}$

$= \lim_{t \to 0} \frac{4}{\cos t (1 - \sqrt{3} \tan t)} = \frac{4}{1(1 - 0)} = 4$

3. 最終的な答え

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

(4) $2$

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

(10) $4$