以下の極限を計算する問題です。 (1) $\lim_{x \to 1} \frac{\sqrt{2x+3} - \sqrt{4x+1}}{x-1}$ (2) $\lim_{x \to \infty} (\sqrt{4x^2+x} - 2x)$ (3) $\lim_{x \to 0} \frac{\sin 3x}{\sin 5x}$ (4) $\lim_{x \to 0} (1+3x)^{\frac{2}{x}}$ (5) $\lim_{x \to 0} \frac{\arctan x - x}{x^3}$

1. 問題の内容

以下の極限を計算する問題です。

(1) $\lim_{x \to 1} \frac{\sqrt{2x+3} - \sqrt{4x+1}}{x-1}$

(2) $\lim_{x \to \infty} (\sqrt{4x^2+x} - 2x)$

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

(4) $\lim_{x \to 0} (1+3x)^{\frac{2}{x}}$

(5) $\lim_{x \to 0} \frac{\arctan x - x}{x^3}$

2. 解き方の手順

(1) $\lim_{x \to 1} \frac{\sqrt{2x+3} - \sqrt{4x+1}}{x-1}$

分母分子に $\sqrt{2x+3} + \sqrt{4x+1}$ をかけます。

$\lim_{x \to 1} \frac{(\sqrt{2x+3} - \sqrt{4x+1})(\sqrt{2x+3} + \sqrt{4x+1})}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{(2x+3) - (4x+1)}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{-2x+2}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{-2(x-1)}{(x-1)(\sqrt{2x+3} + \sqrt{4x+1})} = \lim_{x \to 1} \frac{-2}{\sqrt{2x+3} + \sqrt{4x+1}} = \frac{-2}{\sqrt{5} + \sqrt{5}} = \frac{-2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}$

(2) $\lim_{x \to \infty} (\sqrt{4x^2+x} - 2x)$

分母分子に $\sqrt{4x^2+x} + 2x$ をかけます。

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

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

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

(4) $\lim_{x \to 0} (1+3x)^{\frac{2}{x}}$

$y = (1+3x)^{\frac{2}{x}}$ とおくと、$\ln y = \frac{2}{x} \ln (1+3x)$.

$\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{2 \ln (1+3x)}{x} = 2 \lim_{x \to 0} \frac{\ln (1+3x)}{x} = 2 \lim_{x \to 0} \frac{\frac{3}{1+3x}}{1} = 2 \cdot 3 = 6$.

よって、$\lim_{x \to 0} y = e^6$.

(5) $\lim_{x \to 0} \frac{\arctan x - x}{x^3}$

ロピタルの定理を適用します。

$\lim_{x \to 0} \frac{\arctan x - x}{x^3} = \lim_{x \to 0} \frac{\frac{1}{1+x^2} - 1}{3x^2} = \lim_{x \to 0} \frac{\frac{1 - (1+x^2)}{1+x^2}}{3x^2} = \lim_{x \to 0} \frac{-x^2}{3x^2(1+x^2)} = \lim_{x \to 0} \frac{-1}{3(1+x^2)} = \frac{-1}{3(1+0)} = -\frac{1}{3}$

3. 最終的な答え

(1) $-\frac{1}{\sqrt{5}}$

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

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

(4) $e^6$

(5) $-\frac{1}{3}$