次の6つの極限値を求めます。 (1) $\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}$ (2) $\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}$ (3) $\lim_{x \to \infty} x \sin \frac{1}{4x}$ (4) $\lim_{x \to \pi} \frac{\sin x}{x - \pi}$ (5) $\lim_{x \to 0} \frac{x^2}{1 - \cos 2x}$ (6) $\lim_{x \to \frac{\pi}{2}} \frac{\sin(2 \cos x)}{x - \frac{\pi}{2}}$

はい、承知いたしました。与えられた問題を解きます。

1. 問題の内容

次の6つの極限値を求めます。

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

(2) $\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}$

(3) $\lim_{x \to \infty} x \sin \frac{1}{4x}$

(4) $\lim_{x \to \pi} \frac{\sin x}{x - \pi}$

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

(6) $\lim_{x \to \frac{\pi}{2}} \frac{\sin(2 \cos x)}{x - \frac{\pi}{2}}$

2. 解き方の手順

(1)

$\lim_{x \to 0} \frac{\sin 5x}{\sin 2x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \cdot \frac{5x}{2x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \lim_{x \to 0} \frac{2x}{\sin 2x} \cdot \lim_{x \to 0} \frac{5}{2} = 1 \cdot 1 \cdot \frac{5}{2} = \frac{5}{2}$

(2)

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

(3)

$\lim_{x \to \infty} x \sin \frac{1}{4x}$

$t = \frac{1}{4x}$ と置くと、$x \to \infty$ のとき $t \to 0$

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

(4)

$\lim_{x \to \pi} \frac{\sin x}{x - \pi}$

$t = x - \pi$ と置くと、$x \to \pi$ のとき $t \to 0$

$\lim_{x \to \pi} \frac{\sin x}{x - \pi} = \lim_{t \to 0} \frac{\sin (t + \pi)}{t} = \lim_{t \to 0} \frac{-\sin t}{t} = -1$

(5)

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

(6)

$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2 \cos x)}{x - \frac{\pi}{2}}$

$t = x - \frac{\pi}{2}$ と置くと、$x \to \frac{\pi}{2}$ のとき $t \to 0$

$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2 \cos x)}{x - \frac{\pi}{2}} = \lim_{t \to 0} \frac{\sin(2 \cos (t + \frac{\pi}{2}))}{t} = \lim_{t \to 0} \frac{\sin(2 (-\sin t))}{t} = \lim_{t \to 0} \frac{\sin(-2 \sin t)}{t} = \lim_{t \to 0} \frac{- \sin(2 \sin t)}{t} = \lim_{t \to 0} \frac{- \sin(2 \sin t)}{2 \sin t} \cdot \frac{2 \sin t}{t} = -1 \cdot 2 \cdot 1 = -2$

3. 最終的な答え

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

(2) $2$

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

(4) $-1$

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

(6) $-2$