数学 - 解析学 - 関数の近似、テイラーの定理 - 極限の計算 - 三角関数(正弦)、対数関数、指数関数、不定形の極限、ロピタルの定理

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解析入門(上) (松坂和夫 数学入門シリーズ 4) (松坂 和夫(著)、岩波書店)の第6章(関数の近似、テイラーの定理)、6.2(極限の計算)、問題2の解答を求めてみる。

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

      よって、 ロピタルの 定理より

      $\underset{x\to 0}{\lim }\frac{x-\log \left(1+x\right)}{x^2}=\frac{1}{2}$
   2. 
      
      $\begin{array}{l}\underset{x\to 0}{\lim }\left(\frac{1}{{\sin }^{2}x}-\frac{1}{x^2}\right)\\ =\underset{x\to 0}{\lim }\frac{x^2-{\sin }^{2}x}{x^2{\sin }^{2}x}\\ =\underset{x\to 0}{\lim }\frac{x^2}{{\sin }^{2}x}·\frac{x^2-{\sin }^{2}x}{x^4}\\ \underset{x\to 0}{\lim }\left(x^2-{\sin }^{2}x\right)=0\\ \underset{x\to 0}{\lim }{x}^4=0\\ \frac{d}{\mathrm{dx}}\left(x^2-{\sin }^{2}x\right)=2x-\sin \left(2x\right)\\ \frac{d}{\mathrm{dx}}{x}^4=4x^3\\ \underset{x\to 0}{\lim }\left(2x-\sin \left(2x\right)\right)=0\\ \underset{x\to 0}{\lim }\left(4x^3\right)=0\\ \frac{d}{\mathrm{dx}}\left(2x-\sin \left(2x\right)\right)=2-2\cos \left(2x\right)\\ \frac{d}{\mathrm{dx}}\left(4x^3\right)=12x^2\\ \underset{x\to 0}{\lim }\left(2-2\cos \left(2x\right)\right)=0\\ \underset{x\to 0}{\lim }\left(12x^2\right)=0\\ \frac{d}{\mathrm{dx}}\left(2-2\cos \left(2x\right)\right)=4\sin \left(2x\right)\\ \frac{d}{\mathrm{dx}}\left(12x^2\right)=24x\\ \underset{x\to 0}{\lim }\frac{4\sin \left(2x\right)}{24x}\\ =\frac{1}{3}\underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{2x}\\ =\frac{1}{3}\\ \underset{x\to 0}{\lim }\left(\frac{1}{{\sin }^{2}x}-\frac{1}{x^2}\right)=\frac{1}{3}\end{array}$
   3. 
      
      $\begin{array}{l}\underset{x\to 0}{\lim }\left(x-\sin x\right)=0\\ \underset{x\to 0}{\lim }\left(\tan x-x\right)=0\\ \frac{d}{\mathrm{dx}}\left(x-\sin x\right)=1-\cos x\\ \frac{d}{\mathrm{dx}}\left(\tan x-x\right)\\ =\frac{1}{{\cos }^{2}x}-1\\ =\frac{1-{\cos }^{2}x}{{\cos }^{2}x}\\ =\frac{\left(1+\cos x\right)\left(1-\cos x\right)}{{\cos }^{2}x}\\ \underset{x\to 0}{\lim }\left(1-\cos x\right)·\frac{{\cos }^{2}x}{\left(1+\cos x\right)\left(1-\cos x\right)}=\frac{1}{2}\\ \underset{x\to 0}{\lim }\frac{x-\sin x}{\tan x-x}=\frac{1}{2}\end{array}$
   4. 
      
      $\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\frac{x}{\log x}\left(x^{\frac{1}{x}}-1\right)\right)\\ =\underset{x\to +\infty }{\lim }\frac{x^{\frac{1}{x}}-1}{\log x^{\frac{1}{x}}}\\ \underset{x\to +\infty }{\lim }\left(\log x^{\frac{1}{x}}\right)\\ =\underset{x\to +\infty }{\lim }\frac{\log x}{x}\\ \underset{x\to +0}{\lim }x=+\infty \\ \frac{d}{\mathrm{dx}}\log x=\frac{1}{x}\\ \frac{d}{\mathrm{dx}}x=1\\ \underset{x\to +\infty }{\lim }\frac{1}{x}=0\\ \underset{x\to +\infty }{\lim }\left(\log x^{\frac{1}{x}}\right)=0\\ \underset{x\to +\infty }{\lim }{x}^{\frac{1}{x}}=1\\ \underset{x\to +\infty }{\lim }\left(\frac{x}{\log x}\left(x^{\frac{1}{x}}-1\right)\right)\\ =\underset{y\to 1}{\lim }\frac{y-1}{\log y}\\ \underset{y\to 1}{\lim }\left(y-1\right)=0\\ \underset{y\to 1}{\lim }\log y=0\\ \frac{d}{\mathrm{dy}}\left(y-1\right)=1\\ \frac{d}{\mathrm{dy}}\log y=\frac{1}{y}\\ \underset{y\to 1}{\lim }\frac{1}{\frac{1}{y}}=1\\ \underset{x\to +\infty }{\lim }\left(\frac{x}{\log x}\left(x^{\frac{1}{x}}-1\right)\right)=1\end{array}$
   5. 
      
      $\begin{array}{l}\underset{x\to 0+}{\lim }\log {\left(\frac{1}{\sin x}\right)}^x\\ =\underset{x\to 0+}{\lim }x\log \left(\frac{1}{\sin x}\right)\\ =\underset{x\to 0+}{\lim }\frac{x}{\sin x}\left(\sin x\right)\log \left(\frac{1}{\sin x}\right)\\ =\underset{x\to 0+}{\lim }\frac{x}{\sin x}·\frac{\log \frac{1}{\sin x}}{\frac{1}{\sin x}}\\ \underset{y\to +\infty }{\lim }y=+\infty \\ \frac{d}{\mathrm{dy}}\log y=\frac{1}{y}\\ \frac{d}{\mathrm{dy}}y=1\\ \underset{y\to +\infty }{\lim }\frac{1}{y}=0\\ \underset{y\to +\infty }{\lim }\frac{\log y}{y}=0\\ \underset{x\to 0+}{\lim }\frac{\log \frac{1}{\sin x}}{\frac{1}{\sin x}}=0\\ \underset{x\to 0+}{\lim }\log {\left(\frac{1}{\sin x}\right)}^x=0\\ \underset{x\to 0+}{\lim }{\left(\frac{1}{\sin x}\right)}^x=1\end{array}$
   6. 
      
      $\log \left(1+x\right)=x-\frac{1}{2}{x}^2+\epsilon$

      とおくと、

      $\underset{x\to 0}{\lim }\frac{\epsilon }{x^2}=0$

      が成り立つ。

      よって、

      $\begin{array}{l}\log {\left(1+x\right)}^{\frac{1}{x}}\\ =\frac{1}{x}\log \left(1+x\right)\\ =\frac{1}{x}\left(x-\frac{1}{2}{x}^2+\epsilon \right)\\ =1-\frac{1}{2}x+\frac{\epsilon }{x}\\ \underset{x\to 0}{\lim }\frac{\log {\left(1+x\right)}^{\frac{1}{x}}-1}{x}\\ =\underset{x\to 0}{\lim }\left(-\frac{1}{2}+\frac{\epsilon }{x^2}\right)\\ =-\frac{1}{2}\\ \underset{x\to 0}{\lim }\frac{e-{\left(1+x\right)}^{\frac{1}{x}}}{x}\\ =\underset{x\to 0}{\lim }\frac{\log {\left(1+x\right)}^{\frac{1}{x}}-1}{x}·\frac{e-{\left(1+x\right)}^{\frac{1}{x}}}{\log {\left(1+x\right)}^{\frac{1}{x}}-1}\\ =\underset{x\to 0}{\lim }\frac{\log {\left(1+x\right)}^{\frac{1}{x}}-1}{x}·\frac{e-e^{\left(\log {\left(1+x\right)}^{\frac{1}{x}}\right)}}{\log {\left(1+x\right)}^{\frac{1}{x}}-1}\\ =\underset{x\to 0}{\lim }\frac{\log {\left(1+x\right)}^{\frac{1}{x}}-1}{x}·e·\frac{1-\exp \left(\log {\left(1+x\right)}^{\frac{1}{x}}-1\right)}{\log {\left(1+x\right)}^{\frac{1}{x}}-1}\\ \underset{x\to 0}{\lim }\log {\left(1+x\right)}^{\frac{1}{x}}\\ =\underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}\\ \underset{x\to 0}{\lim }\log \left(1+x\right)=0\\ \underset{x\to 0}{\lim }x=0\\ \frac{d}{\mathrm{dx}}\log \left(1+x\right)=\frac{1}{1+x}\\ \frac{d}{\mathrm{dx}}x=1\\ \underset{x\to 0}{\lim }\frac{1}{1+x}=1\\ \underset{x\to 0}{\lim }\log {\left(1+x\right)}^{\frac{1}{x}}=1\\ \underset{y\to 1}{\lim }\left(1-e^{y-1}\right)=0\\ \underset{y\to 1}{\lim }\left(y-1\right)=0\\ \frac{d}{\mathrm{dy}}\left(1-e^{y-1}\right)=-e^{y-1}\\ \frac{d}{\mathrm{dy}}\left(y-1\right)=-1\\ \underset{y\to 1}{\lim }{e}^{y-1}=-1\\ \underset{y\to 1}{\lim }\frac{1-e^{y-1}}{y-1}=-1\\ \underset{x\to 0}{\lim }\frac{1-\exp \left(\log {\left(1+x\right)}^{\frac{1}{x}}-1\right)}{\log {\left(1+x\right)}^{\frac{1}{x}}-1}=-1\\ \underset{x\to 0}{\lim }\frac{e-{\left(1+x\right)}^{\frac{1}{x}}}{x}=-\frac{1}{2}e\left(-1\right)=\frac{e}{2}\end{array}$