与えられた10個の極限値を計算し、最後に与えられた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個の極限値を計算し、最後に与えられた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) $\lim_{x \to 0} \frac{\tan x - x}{x^3}$

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ...$なので、

$\lim_{x \to 0} \frac{(x + \frac{x^3}{3} + \frac{2x^5}{15} + ...) - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} + \frac{2x^5}{15} + ...}{x^3} = \frac{1}{3}$

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

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

$\sin^2 x = (x - \frac{x^3}{6} + \frac{x^5}{120} - ...)^2 = x^2 - \frac{x^4}{3} + \frac{2x^6}{45} - ...$

$\frac{1}{\sin^2 x} = \frac{1}{x^2 - \frac{x^4}{3} + \frac{2x^6}{45} - ...} = \frac{1}{x^2}(1 - \frac{x^2}{3} + \frac{2x^4}{45} - ...)^{-1} = \frac{1}{x^2}(1 + \frac{x^2}{3} + O(x^4))$

$\lim_{x \to 0} (\frac{1}{x^2} - \frac{1}{\sin^2 x}) = \lim_{x \to 0} (\frac{1}{x^2} - \frac{1}{x^2}(1 + \frac{x^2}{3} + ...)) = \lim_{x \to 0} (-\frac{1}{3} + O(x^2)) = -\frac{1}{3}$

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

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

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

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

$\lim_{x \to 0} \frac{e^x - e^{\sin x}}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} + O(x^4)}{x^3} = \frac{1}{3}$

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

$t = \sin x$ とおくと、$x \to \pi/2$ のとき、$t \to 1$

$\lim_{x \to \pi/2} \frac{e^{\sin x} - e}{\log \sin x} = \lim_{t \to 1} \frac{e^t - e}{\log t} = \lim_{t \to 1} \frac{e^t}{\frac{1}{t}} = \lim_{t \to 1} te^t = e$

(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、$h \to 0$

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

$\lim_{h \to 0} \frac{(1+h)e^{(1+h)\log(1+h)} - (1+h)}{-h + (h - \frac{h^2}{2} + ...)} = \lim_{h \to 0} \frac{(1+h)e^{(1+h)(h-\frac{h^2}{2} + ...)} - (1+h)}{-\frac{h^2}{2}} = \lim_{h \to 0} \frac{(1+h)(1 + (1+h)(h-\frac{h^2}{2} + ...) + ...) - (1+h)}{-\frac{h^2}{2}} = \lim_{h \to 0} \frac{(1+h)(1 + h - \frac{h^2}{2} + h^2 + ...) - (1+h)}{-\frac{h^2}{2}} = \lim_{h \to 0} \frac{1+h+h+h^2+\frac{h^2}{2}-(1+h)}{-\frac{h^2}{2}} = \lim_{h \to 0} \frac{h - \frac{h^2}{2}+h^2+... +O(h^3)}{h - \frac{h^2}{2}+ O(h^3)} = \lim_{h \to 0} \frac{(1+h) - \frac{h^2}{2} + O(h^3)}{(1-h) - (h^2/2) + O(h^3)}$

$= \lim_{h \to 0} \frac{2h+O(h^2)}{-\frac{h^2}{2}} = -2$

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

$(1+x)^{1/x} = e^{\frac{1}{x}\log(1+x)} = e^{\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}-...)} = e^{1-\frac{x}{2}+\frac{x^2}{3}-...} = e \cdot e^{-\frac{x}{2}+\frac{x^2}{3}-...} = e(1 + (-\frac{x}{2}+\frac{x^2}{3}-...) + \frac{(-\frac{x}{2}+\frac{x^2}{3}-...)^2}{2} + ...) = e(1-\frac{x}{2}+\frac{x^2}{3} + \frac{x^2}{8} + ...) = e(1-\frac{x}{2}+\frac{11x^2}{24} + ...)$

$\lim_{x \to 0} \frac{(1+x)^{1/x} - e}{x} = \lim_{x \to 0} \frac{e(1-\frac{x}{2}+\frac{11x^2}{24} + ...) - e}{x} = \lim_{x \to 0} \frac{e(-\frac{x}{2}+\frac{11x^2}{24} + ...)}{x} = -\frac{e}{2}$

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

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

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

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

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

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

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

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

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

$\log(\frac{a^x + b^x}{2}) = \log(1 + x \log \sqrt{ab} + ...) = x \log \sqrt{ab} + O(x^2)$

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

平方根に入るものは $ab$.

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

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

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

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

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

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

展開式: $\frac{1}{1-x} = 1 + x + x^2 + ... + Ax^n + ...$より、$A = 1$

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