1.8 解答

1.8

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

(b) $e^{h} - 1 = x$ とおくと $h = \log{(1+x)}$ で $x \to 0$ と $h \to 0$ は同値．よって 1-8-1aより

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

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

(d)

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

(e) $\displaystyle{\lim_{x \to 0}x^{x} = \lim_{x \to 0}e^{x \log{x}}}$ . ここで $\lim_{x \to 0} x \log{x}$ を求める．

$\displaystyle \lim_{x \to 0} x \log{x} = \lim_{x \to 0} \frac{\log{x}}{\frac{1}{x}}$

より $\log{x} = t$ とおくと $x = e^{t}$ . よって例題1.32より $\displaystyle{\lim_{x \to 0} \frac{\log{x}}{\frac{1}{x}} = \lim_{t \to - \infty} \frac{t}{e^{-t}} = 0}$