与えられた6つの関数の極限値を求めます。 (1) $\lim_{x \to 0} \frac{\sqrt{2+x^2}-\sqrt{2-x^2}}{x^2}$ (2) $\lim_{x \to \infty} \sqrt{x}(\sqrt{2x} - \sqrt{2x+1})$ (3) $\lim_{x \to 0} x\sin(\frac{1}{x})$ (4) $\lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)}$ (5) $\lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2}$ (6) $\lim_{x \to 1} \frac{x}{x-1}$

1. 問題の内容

与えられた6つの関数の極限値を求めます。

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

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

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

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

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

(6) $\lim_{x \to 1} \frac{x}{x-1}$

2. 解き方の手順

(1) 分母分子に$\sqrt{2+x^2} + \sqrt{2-x^2}$をかけます。

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

(2) 分母分子に$\sqrt{2x} + \sqrt{2x+1}$をかけます。

$\lim_{x \to \infty} \sqrt{x}\frac{(\sqrt{2x} - \sqrt{2x+1})(\sqrt{2x} + \sqrt{2x+1})}{\sqrt{2x} + \sqrt{2x+1}} = \lim_{x \to \infty} \sqrt{x}\frac{2x - (2x+1)}{\sqrt{2x} + \sqrt{2x+1}} = \lim_{x \to \infty} \frac{-\sqrt{x}}{\sqrt{2x} + \sqrt{2x+1}} = \lim_{x \to \infty} \frac{-\sqrt{x}}{\sqrt{x}(\sqrt{2} + \sqrt{2+1/x})} = \lim_{x \to \infty} \frac{-1}{\sqrt{2} + \sqrt{2+1/x}} = \frac{-1}{\sqrt{2} + \sqrt{2}} = \frac{-1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}$

(3) $-|x| \le x\sin(\frac{1}{x}) \le |x|$なので、$\lim_{x \to 0} x\sin(\frac{1}{x}) = 0$ （はさみうちの原理）。

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

(5) $e^x = 1 + x + \frac{x^2}{2} + O(x^3)$、$e^{-x} = 1 - x + \frac{x^2}{2} + O(x^3)$より、

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

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

もしくは、ロピタルの定理を使う：

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

(6) $\lim_{x \to 1} \frac{x}{x-1}$

$x \to 1^+$のとき$\frac{x}{x-1} \to \infty$

$x \to 1^-$のとき$\frac{x}{x-1} \to -\infty$

よって極限は存在しません。

3. 最終的な答え

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

(2) $-\frac{\sqrt{2}}{4}$

(3) 0

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

(5) 1

(6) 極限は存在しない