与えられた10個の極限値を求め、最後の展開式における係数Aを求める問題です。

1. 問題の内容

与えられた10個の極限値を求め、最後の展開式における係数Aを求める問題です。

2. 解き方の手順

(1) $\lim_{x \to 0} \frac{\tan x - x}{x^3}$

ロピタルの定理を3回適用します。

$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{2 \sec^2 x \tan x}{6x} = \lim_{x \to 0} \frac{2 \sec^4 x + 4 \sec^2 x \tan^2 x}{6} = \frac{2}{6} = \frac{1}{3}$

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

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

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$を用いると、

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

$\sin^2 x - x^2 = -\frac{x^4}{3} + \dots$

$x^2 \sin^2 x = x^4 + \dots$

$\lim_{x \to 0} \frac{-\frac{x^4}{3}}{x^4} = -\frac{1}{3}$

(3) $\lim_{x \to 0} \frac{e^x - e^{\sin x}}{x^3}$

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$

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

$e^x - e^{\sin x} = \frac{x^3}{6} - (-\frac{x^3}{6}) + \dots = \frac{x^3}{3} + O(x^4)$

$\lim_{x \to 0} \frac{\frac{x^3}{3}}{x^3} = \frac{1}{3}$

(4) $\lim_{x \to \pi/2} \frac{e^{\sin x} - e}{\log(\sin x)}$

$x = \pi/2 + h$とすると$x \to \pi/2$のとき$h \to 0$

$\sin x = \sin(\pi/2 + h) = \cos h$

$\lim_{h \to 0} \frac{e^{\cos h} - e}{\log(\cos h)}$

ロピタルの定理を適用すると、

$\lim_{h \to 0} \frac{-e^{\cos h} \sin h}{-\frac{\sin h}{\cos h}} = \lim_{h \to 0} e^{\cos h} \cos h = e$

(5) $\lim_{x \to 1} \frac{x^x - x}{1 - x + \log x}$

$x = 1 + h$とおくと、$x \to 1$のとき$h \to 0$

$x^x = (1 + h)^{1+h} = e^{(1+h) \log(1+h)} = e^{(1+h)(h - \frac{h^2}{2} + \dots)} = e^{h + \frac{h^2}{2} + O(h^3)} = 1 + (h + \frac{h^2}{2}) + \frac{(h + \frac{h^2}{2})^2}{2} + \dots = 1 + h + \frac{h^2}{2} + \frac{h^2}{2} + \dots = 1 + h + h^2 + \dots$

$x^x - x = 1 + h + h^2 - (1 + h) + \dots = h^2 + \dots$

$1 - x + \log x = 1 - (1 + h) + (h - \frac{h^2}{2} + \dots) = -\frac{h^2}{2} + \dots$

$\lim_{h \to 0} \frac{h^2}{-\frac{h^2}{2}} = -2$

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

$(1+x)^{1/x} = e^{\frac{1}{x} \log(1+x)} = e^{\frac{1}{x} (x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e e^{-\frac{x}{2} + \frac{x^2}{3} - \dots} = e (1 - \frac{x}{2} + \frac{x^2}{3} + \frac{(\frac{-x}{2})^2}{2!} + \dots) = e (1 - \frac{x}{2} + \frac{11x^2}{24} + \dots)$

$(1+x)^{1/x} - e = e (-\frac{x}{2} + \frac{11x^2}{24} + \dots)$

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

(7) $\lim_{x \to \infty} x^{1/x}$

$\lim_{x \to \infty} x^{1/x} = \lim_{x \to \infty} e^{\frac{\log x}{x}}$

$\lim_{x \to \infty} \frac{\log x}{x} = 0$

$\lim_{x \to \infty} e^{\frac{\log x}{x}} = e^0 = 1$

(8) $\lim_{x \to 0} (\frac{a^x + b^x}{2})^{1/x}$

$\lim_{x \to 0} (\frac{a^x + b^x}{2})^{1/x} = \lim_{x \to 0} e^{\frac{1}{x} \log (\frac{a^x + b^x}{2})}$

$a^x = 1 + x \log a + \frac{x^2}{2} (\log a)^2 + \dots$

$b^x = 1 + x \log b + \frac{x^2}{2} (\log b)^2 + \dots$

$\frac{a^x + b^x}{2} = \frac{2 + x(\log a + \log b) + \dots}{2} = 1 + x \frac{\log a + \log b}{2} + \dots$

$\log (\frac{a^x + b^x}{2}) = x \frac{\log a + \log b}{2} + \dots = x \log \sqrt{ab} + \dots$

$\lim_{x \to 0} e^{\frac{x \log \sqrt{ab}}{x}} = e^{\log \sqrt{ab}} = \sqrt{ab}$

平方根に入るものは $ab$

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

$\log (1 + \frac{1}{x}) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \dots$

$x - x^2 \log (1 + \frac{1}{x}) = x - x^2 (\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \dots) = x - x + \frac{1}{2} - \frac{1}{3x} + \dots = \frac{1}{2} - \frac{1}{3x} + \dots$

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

(10) $\lim_{x \to \infty} \log x \log (1 + \frac{1}{x})$

$\lim_{x \to \infty} \log x \log (1 + \frac{1}{x}) = \lim_{x \to \infty} \log x (\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \dots) = \lim_{x \to \infty} \frac{\log x}{x} = 0$

展開式について、

$\frac{1}{1-x} = 1 + x + x^2 + \dots + x^n + \dots$

よって、$A = 1$

3. 最終的な答え

(1) 1/3

(2) -1/3

(3) 1/3

(4) e

(5) -2

(6) -e/2

(7) 1

(8) ab

(9) 1/2

(10) 0

A = 1