解析式 f(x)=(1+x)1/xf(x)=\left( 1+x \right) ^{1/x}f(x)=(1+x)1/x 与 f(x)=(1+1x)xf(x)=\left( 1+\dfrac{1}{x} \right) ^xf(x)=(1+x1)x 图像 极限 limx→0(1+x)1/x=e\lim_{x\to0}\left( 1+x \right) ^{1/x}=elimx→0(1+x)1/x=e limx→+∞(1+x)1/x=1\lim_{x\to+\infty}\left( 1+x \right) ^{1/x}=1limx→+∞(1+x)1/x=1 limx→0+(1+1x)x=1\lim_{x\to0^+}\left( 1+\dfrac{1}{x} \right) ^x=1limx→0+(1+x1)x=1 limx→+∞(1+1x)x=e\lim_{x\to+\infty}\left( 1+\dfrac{1}{x} \right) ^x=elimx→+∞(1+x1)x=e 证明 f(n)=(1+1n)nf(n)=\left( 1+\dfrac{1}{n} \right) ^nf(n)=(1+n1)n 数列递减 f(n)n+1=1(1+1n)(1+1n)⋯(1+1n)n+1<n+1+1n+1=1+1n+1\begin{aligned} \sqrt[n+1]{f(n)}&=\sqrt[n+1]{1\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{n} \right)\cdots\left( 1+\dfrac{1}{n} \right)}\\ &<\dfrac{n+1+1}{n+1}\\ &=1+\dfrac{1}{n+1} \end{aligned}n+1f(n)=n+11(1+n1)(1+n1)⋯(1+n1)<n+1n+1+1=1+n+11 ⟺ f(n)<(1+1n+1)n+1=f(n+1)\iff f(n)<\left( 1+\dfrac{1}{n+1} \right) ^{n+1}=f(n+1)⟺f(n)<(1+n+11)n+1=f(n+1)
