与えられた3つの極限を計算します。 (1) $\lim_{x \to -\infty} (\sqrt{9x^2 + x} + 3x)$ (2) $\lim_{\theta \to 0} \frac{1 - \cos \theta}{4 \theta \sin \theta}$ (3) $\lim_{x \to 0} \frac{\sqrt{1 - \cos 6x}}{x}$

1. 問題の内容

与えられた3つの極限を計算します。

(1) $\lim_{x \to -\infty} (\sqrt{9x^2 + x} + 3x)$

(2) $\lim_{\theta \to 0} \frac{1 - \cos \theta}{4 \theta \sin \theta}$

(3) $\lim_{x \to 0} \frac{\sqrt{1 - \cos 6x}}{x}$

2. 解き方の手順

(1) $\lim_{x \to -\infty} (\sqrt{9x^2 + x} + 3x)$

$x \to -\infty$ なので、$x < 0$ です。

$\sqrt{9x^2 + x} + 3x = \frac{(\sqrt{9x^2 + x} + 3x)(\sqrt{9x^2 + x} - 3x)}{\sqrt{9x^2 + x} - 3x} = \frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} - 3x} = \frac{x}{\sqrt{9x^2 + x} - 3x}$

ここで、$x = -t$ とおくと、$x \to -\infty$ のとき $t \to \infty$ となり、

$\frac{x}{\sqrt{9x^2 + x} - 3x} = \frac{-t}{\sqrt{9t^2 - t} + 3t} = \frac{-t}{t\sqrt{9 - \frac{1}{t}} + 3t} = \frac{-1}{\sqrt{9 - \frac{1}{t}} + 3}$

$\lim_{t \to \infty} \frac{-1}{\sqrt{9 - \frac{1}{t}} + 3} = \frac{-1}{\sqrt{9 - 0} + 3} = \frac{-1}{3 + 3} = -\frac{1}{6}$

(2) $\lim_{\theta \to 0} \frac{1 - \cos \theta}{4 \theta \sin \theta}$

$\lim_{\theta \to 0} \frac{1 - \cos \theta}{4 \theta \sin \theta} = \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} \cdot \frac{\theta}{\sin \theta} \cdot \frac{1}{4} = \lim_{\theta \to 0} \frac{2 \sin^2 \frac{\theta}{2}}{\theta^2} \cdot \frac{\theta}{\sin \theta} \cdot \frac{1}{4} = \lim_{\theta \to 0} 2 \cdot (\frac{\sin \frac{\theta}{2}}{\theta})^2 \cdot \frac{\theta}{\sin \theta} \cdot \frac{1}{4}$

$= 2 \cdot (\frac{1}{2})^2 \cdot 1 \cdot \frac{1}{4} = 2 \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{8}$

または、$1 - \cos \theta = 2 \sin^2(\theta/2)$ なので

$\frac{1 - \cos \theta}{4 \theta \sin \theta} = \frac{2 \sin^2(\theta/2)}{4 \theta \sin \theta} = \frac{\sin^2(\theta/2)}{2 \theta \sin \theta} = \frac{\sin(\theta/2)}{\theta/2} \cdot \frac{\sin(\theta/2)}{\theta/2} \cdot \frac{\theta}{\sin \theta} \cdot \frac{1}{8}$

$\lim_{\theta \to 0} \frac{\sin(\theta/2)}{\theta/2} \cdot \frac{\sin(\theta/2)}{\theta/2} \cdot \frac{\theta}{\sin \theta} \cdot \frac{1}{8} = 1 \cdot 1 \cdot 1 \cdot \frac{1}{8} = \frac{1}{8}$

(3) $\lim_{x \to 0} \frac{\sqrt{1 - \cos 6x}}{x}$

$1 - \cos 6x = 2 \sin^2 3x$

$\lim_{x \to 0} \frac{\sqrt{1 - \cos 6x}}{x} = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 3x}}{x} = \lim_{x \to 0} \frac{\sqrt{2} |\sin 3x|}{x}$

ここで、$x \to +0$ のとき、$\sin 3x > 0$ なので

$\lim_{x \to +0} \frac{\sqrt{2} \sin 3x}{x} = \sqrt{2} \cdot 3 = 3\sqrt{2}$

$x \to -0$ のとき、$\sin 3x < 0$ なので

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

したがって、$\lim_{x \to 0} \frac{\sqrt{1 - \cos 6x}}{x} = 3\sqrt{2}$

3. 最終的な答え

(1) $-\frac{1}{6}$

(2) $\frac{1}{8}$

(3) $3\sqrt{2}$