(10) $\lim_{x \to 0} \frac{1 - \frac{x^2}{2} - \cos x}{\sin^4 x}$ (11) $\lim_{x \to 0} (\frac{1}{x} - \frac{1}{\sin x})$ (12) $\lim_{x \to +0} x^x$

1. 問題の内容

(10) $\lim_{x \to 0} \frac{1 - \frac{x^2}{2} - \cos x}{\sin^4 x}$

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

(12) $\lim_{x \to +0} x^x$

2. 解き方の手順

(10)

まず、$\cos x$ の Taylor 展開を求めます。

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

$1 - \frac{x^2}{2} - \cos x = 1 - \frac{x^2}{2} - (1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + ...)$

$1 - \frac{x^2}{2} - \cos x = -\frac{x^4}{24} + \frac{x^6}{720} - ...$

次に、$\sin x$ の Taylor 展開を求めます。

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$

$\sin^4 x = (x - \frac{x^3}{6} + ...)^4 = x^4(1 - \frac{x^2}{6} + ...)^4 = x^4 - \frac{2x^6}{3} + O(x^8)$

したがって、

$\lim_{x \to 0} \frac{1 - \frac{x^2}{2} - \cos x}{\sin^4 x} = \lim_{x \to 0} \frac{-\frac{x^4}{24} + \frac{x^6}{720} - ...}{x^4 - \frac{2x^6}{3} + ...} = \lim_{x \to 0} \frac{-\frac{1}{24} + \frac{x^2}{720} - ...}{1 - \frac{2x^2}{3} + ...} = -\frac{1}{24}$

(11)

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

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$

$\sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - ...$

$x \sin x = x(x - \frac{x^3}{6} + \frac{x^5}{120} - ...) = x^2 - \frac{x^4}{6} + \frac{x^6}{120} - ...$

$\lim_{x \to 0} \frac{\sin x - x}{x \sin x} = \lim_{x \to 0} \frac{-\frac{x^3}{6} + \frac{x^5}{120} - ...}{x^2 - \frac{x^4}{6} + \frac{x^6}{120} - ...} = \lim_{x \to 0} \frac{-\frac{x}{6} + \frac{x^3}{120} - ...}{1 - \frac{x^2}{6} + \frac{x^4}{120} - ...} = 0$

(12)

$\lim_{x \to +0} x^x$

$y = x^x$

$\ln y = x \ln x$

$\lim_{x \to +0} x \ln x = \lim_{x \to +0} \frac{\ln x}{\frac{1}{x}}$

ロピタルの定理より

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

$\lim_{x \to +0} \ln y = 0$

$\lim_{x \to +0} y = e^0 = 1$

3. 最終的な答え

(10) -1/24

(11) 0

(12) 1