以下の極限を計算します。 (1) $\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1}$ (2) $\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - x)$ (3) $\lim_{x \to \infty} \frac{\sin 3x}{x}$ (4) $\lim_{x \to 0} \frac{\tan^{-1} x}{x}$ (5) $\lim_{x \to 0} (1 + x + x^2)^{\frac{1}{x}}$ (6) $\lim_{x \to 0} \frac{x^2 \sin(\frac{1}{x})}{\sin x}$

1. 問題の内容

以下の極限を計算します。

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

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

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

(4) $\lim_{x \to 0} \frac{\tan^{-1} x}{x}$

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

(6) $\lim_{x \to 0} \frac{x^2 \sin(\frac{1}{x})}{\sin x}$

2. 解き方の手順

(1)

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

(2)

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

(3)

$-\frac{1}{x} \leq \frac{\sin 3x}{x} \leq \frac{1}{x}$.

$\lim_{x \to \infty} \frac{-1}{x} = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$.

By squeeze theorem, $\lim_{x \to \infty} \frac{\sin 3x}{x} = 0$

(4)

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

Alternatively, let $f(x) = \tan^{-1} x$.

$\lim_{x \to 0} \frac{\tan^{-1} x}{x} = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = f'(0)$.

$f'(x) = \frac{1}{1 + x^2}$.

$f'(0) = \frac{1}{1 + 0^2} = 1$.

(5)

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

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

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

(6)

$\lim_{x \to 0} \frac{x^2 \sin(\frac{1}{x})}{\sin x} = \lim_{x \to 0} \frac{x}{\sin x} x \sin(\frac{1}{x})$.

Since $\lim_{x \to 0} \frac{x}{\sin x} = 1$ and $-1 \le \sin(\frac{1}{x}) \le 1$, we have $-|x| \le x \sin(\frac{1}{x}) \le |x|$.

By squeeze theorem, $\lim_{x \to 0} x \sin(\frac{1}{x}) = 0$.

Then, $\lim_{x \to 0} \frac{x^2 \sin(\frac{1}{x})}{\sin x} = (\lim_{x \to 0} \frac{x}{\sin x}) (\lim_{x \to 0} x \sin(\frac{1}{x})) = 1 \times 0 = 0$.

3. 最終的な答え

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

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

(3) $0$

(4) $1$

(5) $e$

(6) $0$