## 2019年10月28日月曜日

### 数学 - 解析学 - 関数の近似、テイラーの定理 - 極限の計算 - 不定形の極限、ロピタルの定理

1. $\begin{array}{l}\underset{x\to +\infty }{\mathrm{lim}}x\left({e}^{\frac{1}{x}}-1\right)\\ =\underset{y\to +0}{\mathrm{lim}}\frac{\left({e}^{y}-1\right)}{y}\\ \frac{d}{\mathrm{dy}}\left({e}^{y}-1\right)={e}^{y}\\ \frac{\mathrm{dy}}{\mathrm{dy}}=1\\ \underset{y\to +0}{\mathrm{lim}}\frac{{e}^{y}}{1}=1\\ \underset{y\to +0}{\mathrm{lim}}\left({e}^{y}-1\right)=0\\ \underset{y\to +0}{\mathrm{lim}}y=0\end{array}$

よって、 ロピタルの定理より

$\underset{x\to +\infty }{\mathrm{lim}}x\left({e}^{\frac{1}{x}}-1\right)=1$

2. $\begin{array}{l}\mathrm{log}{x}^{\frac{1}{x}}\\ =\frac{1}{x}\mathrm{log}x\\ \underset{x\to +\infty }{\mathrm{lim}}\frac{1}{x}\mathrm{log}x\\ =\underset{y\to +0}{\mathrm{lim}}\frac{\mathrm{log}{y}^{-1}}{\frac{1}{y}}\\ =-\underset{y\to +0}{\mathrm{lim}}\frac{\mathrm{log}y}{\frac{1}{y}}\\ \underset{y\to +0}{\mathrm{lim}}\frac{1}{y}=\infty \\ \frac{d}{\mathrm{dy}}\mathrm{log}y=\frac{1}{y}\\ \frac{d}{\mathrm{dy}}\frac{1}{y}=\frac{-1}{{y}^{2}}\\ \underset{y\to +0}{\mathrm{lim}}\frac{\frac{1}{y}}{-\frac{1}{{y}^{2}}}=0\end{array}$

よってロピタルの定理より、

$\underset{x\to +\infty }{\mathrm{lim}}\mathrm{log}{x}^{\frac{1}{x}}=0$

ゆえに、

$\underset{x\to +\infty }{\mathrm{lim}}{x}^{\frac{1}{x}}=1$

3. $\begin{array}{l}\underset{x\to 0}{\mathrm{lim}}\left({a}^{x}-{b}^{x}\right)=0\\ \underset{x\to 0}{\mathrm{lim}}x=0\\ \frac{d}{\mathrm{dx}}\left({a}^{x}-{b}^{x}\right)\\ ={a}^{x}\mathrm{log}a-{b}^{x}\mathrm{log}b\\ \frac{d}{\mathrm{dx}}x=1\\ \underset{x\to 0}{\mathrm{lim}}\frac{{a}^{x}\mathrm{log}a-{b}^{x}\mathrm{log}b}{1}\\ =\mathrm{log}a-\mathrm{log}b\\ =\mathrm{log}\frac{a}{b}\\ \underset{x\to 0}{\mathrm{lim}}\frac{{a}^{x}-{b}^{x}}{x}=\mathrm{log}\frac{a}{b}\end{array}$

4. $\begin{array}{l}\underset{x\to 0}{\mathrm{lim}}\left(x-\mathrm{arcsin}x\right)=0\\ \underset{x\to 0}{\mathrm{lim}}{x}^{3}=0\\ \frac{d}{\mathrm{dx}}\left(x-\mathrm{arcsin}x\right)=1-\frac{1}{\sqrt{1-{x}^{2}}}\\ \frac{d}{\mathrm{dx}}{x}^{3}=3{x}^{2}\\ \underset{x\to 0}{\mathrm{lim}}\left(1-\frac{1}{\sqrt{1-{x}^{2}}}\right)=0\\ \underset{x\to 0}{\mathrm{lim}}3{x}^{2}=0\\ \frac{d}{\mathrm{dx}}\left(1-\frac{1}{\sqrt{1-{x}^{2}}}\right)\\ =-\frac{x}{\left(1-{x}^{2}\right)\sqrt{1-{x}^{2}}}\\ \frac{d}{\mathrm{dx}}3{x}^{2}=6x\\ \underset{x\to 0}{\mathrm{lim}}\left(-\frac{x}{\left(1-{x}^{2}\right)\sqrt{1-{x}^{2}}}\right)=0\\ \underset{x\to 0}{\mathrm{lim}}6x=0\\ \underset{x\to 0}{\mathrm{lim}}\left(-\frac{x}{{\left(1-x\right)}^{2}\sqrt{1-{x}^{2}}}\right)·\frac{1}{6x}=-\frac{1}{6}\\ \underset{x\to 0}{\mathrm{lim}}\frac{x-\mathrm{arcsin}x}{{x}^{3}}=-\frac{1}{6}\end{array}$

5. $\begin{array}{l}\mathrm{tan}x-\frac{1}{\mathrm{cos}x}\\ =\frac{\mathrm{sin}x-1}{\mathrm{cos}x}\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\left(\mathrm{sin}x-1\right)=0\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\mathrm{cos}x=0\\ \frac{d}{\mathrm{dx}}\left(\mathrm{sin}x-1\right)=\mathrm{cos}x\\ \frac{d}{\mathrm{dx}}\mathrm{cos}x=-\mathrm{sin}x\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\frac{\mathrm{cos}x}{-\mathrm{sin}x}\\ =\underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\left(-\frac{1}{\mathrm{tan}x}\right)\\ =0\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\left(\mathrm{tan}x-\frac{1}{\mathrm{cos}x}\right)=0\end{array}$

6. $\begin{array}{l}\underset{x\to \frac{\pi }{2}}{\mathrm{lim}}x\mathrm{sin}x-\frac{\pi }{2}=0\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\mathrm{cos}x=0\\ \frac{d}{\mathrm{dx}}\left(x\mathrm{sin}x-\frac{\pi }{2}\right)=\mathrm{sin}x-x\mathrm{cos}x\\ \frac{d}{\mathrm{dx}}\mathrm{cos}x=-\mathrm{sin}x\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\frac{\mathrm{sin}x-x\mathrm{cos}x}{-\mathrm{sin}x}=-1\\ \underset{x\to \frac{\pi }{2}}{\mathrm{lim}}\frac{x\mathrm{sin}x-\frac{\pi }{2}}{\mathrm{cos}x}=-1\end{array}$