2.6 不定形の極限値

(a) $\lim_{x \to 0+}\frac{\sin{x}}{\sqrt{x}} = (\frac{0}{0})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to 0+}\frac{\sin{x}}{\sqrt{x}}$	$\displaystyle \overbrace{=}^{*}$	$\displaystyle \lim_{x \to 0+}\frac{\cos{x}}{\frac{1}{2\sqrt{x}}}$
	$\displaystyle =$	$\displaystyle \lim_{x \to 0+}2\sqrt{x}\cos{x} = 0$

(b) $\lim_{x \to 1}\frac{\log{x}}{1 - x} = (\frac{0}{0})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to 1}\frac{\log{x}}{1 - x}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to 1}\frac{\frac{1}{x}}{-1} = -1$

$\displaystyle \lim_{x \to 4}\frac{\sqrt{x} - 2}{x - 4}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to 4}\frac{1}{2\sqrt{x}} = \frac{1}{4}$

(d) $\lim_{x \to 0}\frac{2^x - 1}{x} = (\frac{0}{0})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to 0}\frac{2^x -1}{x}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to 0}2^{x} \log{x} = \log{2}$

(e) $\lim_{x \to 0}\frac{1 - \cos{x}}{3x} = (\frac{0}{0})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to 0}\frac{1 - \cos{x}}{3x}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to 0}\frac{\sin{x}}{3} = 0$

(f) $\lim_{x \to \infty}\frac{x - 1}{x + 1} = (\frac{\infty}{\infty})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to \infty}\frac{x -1}{x+1}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to \infty}\frac{1}{1} = 1$

(g) $\lim_{x \to \infty}\frac{2\sin{x}}{x} = 0$

(h) $\lim_{x \to 0}\frac{e^x - e^{-x}}{x} = (\frac{0}{0})$ 不定形．そこで，L'Hospitalの定理を用いると，

$\displaystyle \lim_{x \to 0}\frac{e^x - e^{-x}}{x}$

$\displaystyle \overbrace{=}^{*}$

$\displaystyle \lim_{x \to 0}(e^{x} + e^{-x}) = 2$