\(\lim_{x \to 0} \frac{\sqrt[n]{1+x} -1}{\frac{x}{n} } =1\)的证明
\[\lim_{x \to 0} \frac{\sqrt[n]{1+x} -1}{\frac{x}{n} }
=\lim_{x \to 0} \frac{\left ( 1+x \right )^{\frac{1}{n} }-1 }{\frac{x}{n} }
=\lim_{x \to 0} \frac{e^{x\frac{1}{n} }-1}{\frac{x}{n} }
=\lim_{x \to 0} \frac{\left ( 1+x\frac{1}{n} \right )^{\frac{1}{x\frac{1}{n} }\times x\frac{1}{n} } -1 }{\frac{x}{n} }
=\lim_{x \to 0} \frac{\frac{x}{n} }{\frac{x}{n} }
=1
\]
e ~ \((1+x)^{\frac{1}{x}}\)---(x-->0)
\(e^{x}\) ~ \(1+x\)---(x-->0)
\(e^{xa}\) ~ \(1+xa\)---(x-->0)
\((1+x)^{\frac{1}{x}xa}\) ~ \(xa\)
\((1+x)^{a}\) ~ {xa}
令a = 1/n
\(\sqrt[n]{1+x}\) ~ \(\frac{x}{n}\)