2.8 解答

2.8

(a) $\displaystyle{f^{(n)}(x) = \cos{(x + \frac{n\pi}{2})}}$ より

$\displaystyle \vert R_{n}\vert = \vert\frac{f^{(n)}(\theta x)}{n!}x^{n}\vert \l... ...rt^{n} \leq \frac{1}{n!}\vert x\vert^{n} \rightarrow 0 (n \rightarrow \infty)$

(b) $\displaystyle{f^{(n)}(x) = (-1)^{n-1} (1 + x)^{-n}}$ より

$\displaystyle \vert R_{n}\vert = \vert\frac{f^{(n)}(\theta x)}{n!}x^{n}\vert \l... ...\theta x)^{-n}}{n!}\vert\vert x\vert^{n} \rightarrow 0 (n \rightarrow \infty)$

$\displaystyle \vert R_{n}\vert = \vert\frac{f^{(n)}(\theta x)}{n!}x^{n}\vert \l... ...x)^{\alpha -n}}{n!}\vert\vert x\vert^{n} \rightarrow 0 (n \rightarrow \infty)$

(a)

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

(b)

$\displaystyle{\lim_{x \to 0}\frac{x - \sin{x}}{x^3} = \lim_{x \to 0}\frac{x - x + \frac{x^3}{6} + o(x^{3})}{x^3} = \frac{1}{6}}$

(c)

$\displaystyle{\lim_{x \to 0}\frac{e^x - 1 - x}{x^2} = \lim_{x \to 0}\frac{1 + x + \frac{x^2}{2} + o(x^2)}{x^2} = \frac{1}{2}}$

(d)

$\displaystyle{\lim_{x \to 0}\frac{x^{\alpha}}{e^x} = \lim_{x \to 0}\frac{x^{\alpha}}{1+ x + \frac{x^2}{2} + \cdots } = 0}$