次の極限値を求めます。 (a) $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ (b) $\lim_{x \to 1} \frac{x^2 - 10x + 9}{x - 1}$ (c) $\lim_{x \to 3} \frac{x^2 - 2x - 3}{x^2 - x - 6}$ (d) $\lim_{x \to 1} \frac{x^8 - 1}{x - 1}$ (e) $\lim_{x \to 2} \frac{x^3 - 2x^2 + x - 2}{2x^2 - x - 6}$ (f) $\lim_{x \to 3} \frac{x - 3}{\sqrt{x + 1} - 2}$ (g) $\lim_{x \to 3} \frac{x - 3}{x - \sqrt{x + 6}}$ (h) $\lim_{x \to 0} \frac{\sin 3x}{2x}$ (i) $\lim_{x \to 0} \frac{\tan 2x}{3x}$

1. 問題の内容

次の極限値を求めます。

(a) $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

(b) $\lim_{x \to 1} \frac{x^2 - 10x + 9}{x - 1}$

(c) $\lim_{x \to 3} \frac{x^2 - 2x - 3}{x^2 - x - 6}$

(d) $\lim_{x \to 1} \frac{x^8 - 1}{x - 1}$

(e) $\lim_{x \to 2} \frac{x^3 - 2x^2 + x - 2}{2x^2 - x - 6}$

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

(g) $\lim_{x \to 3} \frac{x - 3}{x - \sqrt{x + 6}}$

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

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

2. 解き方の手順

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

(b) $\lim_{x \to 1} \frac{x^2 - 10x + 9}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x - 9)}{x - 1} = \lim_{x \to 1} (x - 9) = 1 - 9 = -8$

(c) $\lim_{x \to 3} \frac{x^2 - 2x - 3}{x^2 - x - 6} = \lim_{x \to 3} \frac{(x - 3)(x + 1)}{(x - 3)(x + 2)} = \lim_{x \to 3} \frac{x + 1}{x + 2} = \frac{3 + 1}{3 + 2} = \frac{4}{5}$

(d) $\lim_{x \to 1} \frac{x^8 - 1}{x - 1}$

これは $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n - 1}$ の公式を利用できます。$a = 1, n = 8$ なので、

$\lim_{x \to 1} \frac{x^8 - 1}{x - 1} = 8 \cdot 1^{8 - 1} = 8$

またはロピタルの定理を使用すると、$\lim_{x \to 1} \frac{8x^7}{1} = 8$

(e) $\lim_{x \to 2} \frac{x^3 - 2x^2 + x - 2}{2x^2 - x - 6} = \lim_{x \to 2} \frac{x^2(x - 2) + (x - 2)}{(2x + 3)(x - 2)} = \lim_{x \to 2} \frac{(x^2 + 1)(x - 2)}{(2x + 3)(x - 2)} = \lim_{x \to 2} \frac{x^2 + 1}{2x + 3} = \frac{2^2 + 1}{2 \cdot 2 + 3} = \frac{5}{7}$

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

(g) $\lim_{x \to 3} \frac{x - 3}{x - \sqrt{x + 6}} = \lim_{x \to 3} \frac{(x - 3)(x + \sqrt{x + 6})}{(x - \sqrt{x + 6})(x + \sqrt{x + 6})} = \lim_{x \to 3} \frac{(x - 3)(x + \sqrt{x + 6})}{x^2 - (x + 6)} = \lim_{x \to 3} \frac{(x - 3)(x + \sqrt{x + 6})}{x^2 - x - 6} = \lim_{x \to 3} \frac{(x - 3)(x + \sqrt{x + 6})}{(x - 3)(x + 2)} = \lim_{x \to 3} \frac{x + \sqrt{x + 6}}{x + 2} = \frac{3 + \sqrt{3 + 6}}{3 + 2} = \frac{3 + 3}{5} = \frac{6}{5}$

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

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

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

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

3. 最終的な答え

(a) 4

(b) -8

(c) $\frac{4}{5}$

(d) 8

(e) $\frac{5}{7}$

(f) 4

(g) $\frac{6}{5}$

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

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