次の極限を求める問題です。 (1) $\lim_{x \to 0} \frac{\sin x}{2x}$ (2) $\lim_{x \to 0} \frac{\sin 3x}{x}$ (4) $\lim_{x \to 0} \frac{x}{\cos x}$ (5) $\lim_{x \to 0} \frac{\sin 3x}{\sin 2x}$ (7) $\lim_{x \to 0} \frac{\sin x^2}{x}$ (8) $\lim_{x \to 0} \frac{1 - \cos x}{\sin x}$

1. 問題の内容

次の極限を求める問題です。

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

(2) $\lim_{x \to 0} \frac{\sin 3x}{x}$

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

(5) $\lim_{x \to 0} \frac{\sin 3x}{\sin 2x}$

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

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

2. 解き方の手順

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

$\lim_{x \to 0} \frac{\sin x}{x} = 1$ より、

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

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

$\lim_{x \to 0} \frac{\sin 3x}{3x} = 1$ より、

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

(4) $\lim_{x \to 0} \frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0$

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

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

$\lim_{x \to 0} \frac{\sin x^2}{x^2} = 1$ より、

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

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

$\lim_{x \to 0} \frac{\sin x}{1 + \cos x} = \frac{\sin 0}{1 + \cos 0} = \frac{0}{1 + 1} = \frac{0}{2} = 0$

3. 最終的な答え

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

(2) $3$

(4) $0$

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

(7) $0$

(8) $0$