数学微积分，学习笔记，等价无穷小的证明：(1+x)^a-1 ~ 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
$\sqrt[n]{1+x}$ ~ $\frac{x}{n}$