【微分積分】2-3-1 指数関数・対数関数｜問題集

\(\displaystyle =\lim_{n\to\infty}\left(\frac{n-1}{n}\right)^{-n}\)
\(\displaystyle =\lim_{n\to\infty}\left(\frac{n}{n-1}\right)^n\)
\(\displaystyle =\lim_{n\to\infty}\left(1+\frac{1}{n-1}\right)^n\)
\(\displaystyle =\lim_{n\to\infty}\left(1+\frac{1}{n-1}\right)^{n-1}\left(1+\frac{1}{n-1}\right)\)
\(=e・1\)
\(=e\)

\(\displaystyle =\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n\)
\(=e^{-1}・e\)
\(=1\)

\(\displaystyle =\lim_{x\to0}\left(\frac{e^x-1}{x}+\frac{e^{-x}-1}{-x}\right)\)
\(=1+1\)
\(=2\)

\(\displaystyle =\lim_{x\to\pm\infty}\left\{\left(1+\frac{a}{x}\right)^{\frac{x}{a}}\right\}^a\)
\(=e^a\)

\(\displaystyle =\lim_{x\to0}\log(1+x)^{\frac{1}{x}}\)
\(=\log e\)
\(=1\)

\(x=e^h-1\)とおくと、\(h\to0\)のとき\(x\to0\)
\(\displaystyle =\lim_{x\to0}\frac{x}{\log(1+x)}\)
\(=1\)

\(h=x-a\)とおくと、\(x\to a\)のとき\(h\to0\)
\(\displaystyle =\lim_{h\to0}\frac{\log(\frac{a+h}{a})}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\log(1+\frac{h}{a})}{h}\)
\(\displaystyle =\lim_{h\to0}\log\left(1+\frac{h}{a}\right)^{\frac{1}{h}}\)
\(\displaystyle =\frac{1}{a}\)

\(\displaystyle =\lim_{x\to0}\frac{e^{\alpha\log(1+x)}-1}{x}\)
\(\displaystyle =\lim_{x\to0}\frac{e^{\alpha\log(1+x)}-1}{\alpha\log(1+x)}・\frac{\alpha\log(1+x)}{x}\)
\(\displaystyle =\lim_{h\to0}\frac{e^h-1}{h}\lim_{x\to0}\frac{\alpha\log(1+x)}{x}\alpha\)
\(=1・\alpha\)
\(=\alpha\)

\(\displaystyle =\lim_{x\to0}e^{x\log x}\)
ここで、
\(\displaystyle \lim_{x\to0}x\log x\)
\(\displaystyle =\lim_{x\to0}\frac{\log x}{\frac{1}{x}}\)
\(\displaystyle =\lim_{t\to-\infty}\frac{t}{e^{-t}}\)
\(=0\)
よって、
\(\displaystyle \lim_{x\to0}x^x=1\)

\(\displaystyle \lim_{x\to+0}f(x)=\lim_{x\to+0}\frac{1}{1+e^{\frac{1}{x}}}=\lim_{t\to\infty}\frac{1}{1+e^t}=0\)
\(\displaystyle \lim_{x\to-0}f(x)=\lim_{x\to-0}\frac{1}{1+e^{\frac{1}{x}}}=\lim_{t\to-\infty}\frac{1}{1+e^t}=1\)
また、\(f(0)=0\)より、
\(\displaystyle \lim_{x\to-0}f(x)\neq f(0)\)
よって、\(f(x)\)は\(x=0\)で不連続