与えられた10個の極限値を求める問題と、1つの展開式の係数を求める問題です。

1. 問題の内容

与えられた10個の極限値を求める問題と、1つの展開式の係数を求める問題です。

2. 解き方の手順

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

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ を使うと、

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

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

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

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

$\frac{1}{\sin^2 x} = \frac{1}{x^2 - \frac{x^4}{3} + \frac{2x^6}{45} - \dots} = \frac{1}{x^2} \cdot \frac{1}{1 - \frac{x^2}{3} + \frac{2x^4}{45} - \dots} = \frac{1}{x^2} (1 + \frac{x^2}{3} + \frac{x^4}{15} + \dots) = \frac{1}{x^2} + \frac{1}{3} + \frac{x^2}{15} + \dots$

$\lim_{x \to 0} \left( \frac{1}{x^2} - (\frac{1}{x^2} + \frac{1}{3} + \frac{x^2}{15} + \dots) \right) = \lim_{x \to 0} \left( -\frac{1}{3} - \frac{x^2}{15} - \dots \right) = -\frac{1}{3}$

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

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

$\sin x = x - \frac{x^3}{6} + \dots$

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

$\lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots) - (1 + x + \frac{x^2}{2} - \frac{x^4}{6} + \dots)}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} + O(x^4)}{x^3} = \frac{1}{6}$

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

$y = \sin x$ と置くと、$x \to \pi/2$ のとき $y \to 1$

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

$y = 1 + h$ と置くと、$y \to 1$ のとき $h \to 0$

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

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

$x = 1 + h$ と置くと、$x \to 1$ のとき $h \to 0$

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

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

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

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

$\lim_{x \to 0} (1+x)^{1/x} = e$, $(1+x)^{1/x} = e^{\frac{1}{x} \log(1+x)} = e^{\frac{1}{x}(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots} = e (1 - \frac{x}{2} + \frac{x^2}{3} + \frac{(-\frac{x}{2})^2}{2} + \dots) = e(1 - \frac{x}{2} + \frac{11x^2}{24} + \dots)$

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

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

$y = x^{1/x}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. 最終的な答え

(1) 1/3

(2) -1/3

(3) 1/6

(4) e

(5) -2

(6) -e/2

(7) 1

(8) ab

(9) 1/2

(10) 0

A:1