画像に記載された数学の問題を解きます。 1. 極限値を求める問題が6問

1. 問題の内容

画像に記載された数学の問題を解きます。

1. 極限値を求める問題が6問

2. 関数が $x=0$ で連続かどうかを調べる問題が2問

3. 逆三角関数の値を求める問題が3問

4. 逆三角関数を含む方程式を解く問題が2問

2. 解き方の手順

1. (1)

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

分子を有理化します。

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

1. (2)

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

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

1. (3)

$\lim_{x \to 0} x\sin{\frac{1}{x}}$

$-1 \leq \sin{\frac{1}{x}} \leq 1$

$-|x| \leq x\sin{\frac{1}{x}} \leq |x|$

$\lim_{x \to 0} -|x| = 0$

$\lim_{x \to 0} |x| = 0$

挟み撃ちの原理より

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

1. (4)

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

1. (5)

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

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

またはロピタルの定理より

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

1. (6)

$\lim_{x \to 1} x^{\frac{1}{1-x}}$

$y = x^{\frac{1}{1-x}}$

$\log{y} = \frac{1}{1-x}\log{x}$

$\lim_{x \to 1} \log{y} = \lim_{x \to 1} \frac{\log{x}}{1-x}$

ロピタルの定理より

$\lim_{x \to 1} \frac{\log{x}}{1-x} = \lim_{x \to 1} \frac{\frac{1}{x}}{-1} = -1$

$\lim_{x \to 1} \log{y} = -1$

$\lim_{x \to 1} y = e^{-1} = \frac{1}{e}$

2. (1)

$f(x) = \begin{cases} \frac{\sin{2x}}{x} & (x \neq 0) \\ 1 & (x = 0) \end{cases}$

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

$f(0) = 1$

$\lim_{x \to 0} f(x) \neq f(0)$

よって、x=0で不連続

3. (2)

$f(x) = \begin{cases} 2x\sin{\frac{1}{x}} & (x \neq 0) \\ 1 & (x = 0) \end{cases}$

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

$f(0) = 1$

$\lim_{x \to 0} f(x) \neq f(0)$

よって、x=0で不連続

4. (1)

$\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}$

5. (2)

$\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$

6. (3)

$\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$

7. (1)

$\sin^{-1}x = \tan^{-1}\sqrt{5}$

$\theta = \tan^{-1}\sqrt{5}$

$\tan\theta = \sqrt{5}$

$\sin\theta = \frac{\sqrt{5}}{\sqrt{6}}$

$x = \frac{\sqrt{5}}{\sqrt{6}} = \frac{\sqrt{30}}{6}$

8. (2)

$\sin^{-1}x = \sin^{-1}\frac{1}{3} + \sin^{-1}\frac{7}{9}$

$\sin^{-1}x = \alpha + \beta$

$\alpha = \sin^{-1}\frac{1}{3}, \beta = \sin^{-1}\frac{7}{9}$

$\sin\alpha = \frac{1}{3}, \sin\beta = \frac{7}{9}$

$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}$

$\cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \frac{49}{81}} = \sqrt{\frac{32}{81}} = \frac{4\sqrt{2}}{9}$

$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{1}{3}\cdot\frac{4\sqrt{2}}{9} + \frac{2\sqrt{2}}{3}\cdot\frac{7}{9} = \frac{4\sqrt{2}}{27} + \frac{14\sqrt{2}}{27} = \frac{18\sqrt{2}}{27} = \frac{2\sqrt{2}}{3}$

$x = \frac{2\sqrt{2}}{3}$

3. 最終的な答え

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

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

3. (3): $0$

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

5. (5): $0$

6. (6): $\frac{1}{e}$

7. (1): 不連続

8. (2): 不連続

9. (1): $\frac{\pi}{3}$

1

0. (2): $\frac{2\pi}{3}$

1

1. (3): $\frac{\pi}{6}$

1

2. (1): $\frac{\sqrt{30}}{6}$

1

3. (2): $\frac{2\sqrt{2}}{3}$