与えられた12個の極限値を求める問題です。

1. 問題の内容

与えられた12個の極限値を求める問題です。

2. 解き方の手順

(1) $\lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6}$

$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

$x^2 + x - 6 = (x - 2)(x + 3)$

$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 3} = \frac{2^2 + 2 \cdot 2 + 4}{2 + 3} = \frac{12}{5}$

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

$\lim_{x \to 0} \frac{1}{x} \left( \frac{(x+2)^2 - 4}{(x+2)^2} \right) = \lim_{x \to 0} \frac{x^2 + 4x + 4 - 4}{x(x+2)^2} = \lim_{x \to 0} \frac{x^2 + 4x}{x(x+2)^2} = \lim_{x \to 0} \frac{x(x+4)}{x(x+2)^2} = \lim_{x \to 0} \frac{x+4}{(x+2)^2} = \frac{0+4}{(0+2)^2} = \frac{4}{4} = 1$

(3) $\lim_{x \to 0} \frac{x}{\sqrt{x+16} - 4}$

$\lim_{x \to 0} \frac{x(\sqrt{x+16} + 4)}{(\sqrt{x+16} - 4)(\sqrt{x+16} + 4)} = \lim_{x \to 0} \frac{x(\sqrt{x+16} + 4)}{x+16 - 16} = \lim_{x \to 0} \frac{x(\sqrt{x+16} + 4)}{x} = \lim_{x \to 0} \sqrt{x+16} + 4 = \sqrt{0+16} + 4 = 4 + 4 = 8$

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

$\lim_{x \to \infty} \frac{2 + \frac{1}{x}}{1 + \frac{3}{x}} = \frac{2 + 0}{1 + 0} = 2$

(5) $\lim_{x \to \infty} \frac{3x^2 - 5x - 2}{x^2 - 3x + 2}$

$\lim_{x \to \infty} \frac{3 - \frac{5}{x} - \frac{2}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}} = \frac{3 - 0 - 0}{1 - 0 + 0} = 3$

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

$\lim_{x \to \infty} \frac{2x - \frac{5}{x} + \frac{7}{x^2}}{1 + \frac{2}{x} - \frac{1}{x^2}} = \infty$

(7) $\lim_{x \to \infty} (7^x - 2 \cdot 3^x)$

これは問題が不明確なため、保留とします。

(8) $\lim_{x \to \infty} \log_3 \frac{9x+2}{x+3}$

$\lim_{x \to \infty} \log_3 \frac{9 + \frac{2}{x}}{1 + \frac{3}{x}} = \log_3 \frac{9+0}{1+0} = \log_3 9 = 2$

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

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

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

$\lim_{x \to 0} \frac{\sin 2x}{2x} \cdot \frac{7x}{\sin 7x} \cdot \frac{2x}{7x} = 1 \cdot 1 \cdot \frac{2}{7} = \frac{2}{7}$

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

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

(12) $\lim_{x \to 0} \frac{\sin x - \tan 2x}{x}$

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

3. 最終的な答え

(1) $\frac{12}{5}$

(2) $1$

(3) $8$

(4) $2$

(5) $3$

(6) $\infty$

(7) 回答不能

(8) $2$

(9) $5$

(10) $\frac{2}{7}$

(11) $2$

(12) $-1$