函数的等价无穷小

发布时间:2018-08-10 作者:大扑棱蛾子 阅读次数:
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$$
\text{如果} lim\frac{\beta}{\alpha}=1, \text{那么就说}\beta \text{与} \alpha \text{是等价无穷小,记作} \alpha \sim \beta
$$

几个常用的等价无穷小关系式

$$
\begin{array}{rr}
ln(1+x) \sim x \qquad (x \to 0) \
e^x-1 \sim x \qquad (x \to 0) \
(1+x)^\alpha -1 \sim \alpha x \qquad (x \to 0)
\end{array}
$$

证明过程

$$\text{例1:求}\quad \lim\nolimits_{x \to 0} \frac{\log_a(1+x)}{x}$$

$$
\begin{array}{cc}
\displaystyle
\lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0}\log_a(1+x)^\frac{1}{x} \\
\text{令} \quad f(x)=\log_ax, \quad g(x)=(1+x)^\frac{1}{x} \\
\displaystyle
\lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0} f \circ g=
f(\lim\limits_{x \to 0}g(x))=f(\lim\limits_{x \to 0}(1+x)^\frac{1}{x})\\
\text{令} a = \frac{1}{x} \text{,那么} x \to 0 \iff a \to \infty \text{所以} \\
\lim\limits_{x \to 0}(1+x)^\frac{1}{x}=\lim\limits_{a \to \infty}(1+\frac{1}{a})^{a}=e \\
\lim\limits_{x \to 0} f \circ g = \log_ae \\
\displaystyle
\lim\limits_{x \to 0}\frac{\log_a(1+x)}{x}=\lim\limits_{x \to 0}\log_ae=\lim\limits_{x \to 0}\frac{1}{\log_ea}=\lim\limits_{x \to 0}\frac{1}{\ln:a}
\end{array}
$$

将上面证明过程中的$a$换成$e$就有
$$
\lim\limits_{x \to 0}\frac{\ln(1+x)}{x}=\frac{1}{\ln:e}=1
$$


$$\text{例2:求}\quad \lim\nolimits_{x \to 0} \frac{a^x-1}{x}$$

$$
Let \quad a^x-1=t, so \quad x=\log_a(1+t),When \quad x \to 0 \quad then \quad t \to 0,\quad So \quad that
$$

$$
\begin{array}{ll}
\displaystyle
\lim\limits_{x \to 0} \frac{a^x-1}{x} &\displaystyle = \lim\limits_{t \to 0} \frac{t}{\log_a(1+t)} \\
&\displaystyle = \lim\limits_{t \to 0}: t \cdot \frac{1}{\log_a(1+t)} = \lim\limits_{t \to 0}: t \cdot \log_{(1+t)}a \\
&\displaystyle = \lim\limits_{t \to 0} : \log_{(1+t)^{\frac{1}{t}}}a = \log_{(\lim\nolimits_{x \to 0}(1+t)^{\frac{1}{t}})}a\\
&\displaystyle = \log_e a = \ln a
\end{array}
$$

将上面证明过程中的$$a$$换成$$e$$就有
$$
\lim\limits_{x \to 0} \frac{e^x-1}{x} = \ln e=1
$$

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