数学微积分，学习笔记，等价无穷小的证明：(1+x)^a-1 ~ ax

标签：frac -- lim 微积分 xa sqrt --- 无穷小 ax

\(\lim_{x \to 0} \frac{\sqrt[n]{1+x} -1}{\frac{x}{n} } =1\)的证明

\[\lim_{x \to 0} \frac{\sqrt[n]{1+x} -1}{\frac{x}{n} } =\lim_{x \to 0} \frac{\left ( 1+x \right )^{\frac{1}{n} }-1 }{\frac{x}{n} } =\lim_{x \to 0} \frac{e^{x\frac{1}{n} }-1}{\frac{x}{n} } =\lim_{x \to 0} \frac{\left ( 1+x\frac{1}{n} \right )^{\frac{1}{x\frac{1}{n} }\times x\frac{1}{n} } -1 }{\frac{x}{n} } =\lim_{x \to 0} \frac{\frac{x}{n} }{\frac{x}{n} } =1 \]

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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
\(\sqrt[n]{1+x}\) ~ \(\frac{x}{n}\)

标签：frac,--,lim,微积分,xa,sqrt,---,无穷小,ax
From： https://www.cnblogs.com/l25428455/p/17826937.html