常用等价无穷小
重要极限罗列¶
\[
\begin{array}{lll}
\sin x \sim x&1 - \cos x \sim \frac{1}{2}x^2&\tan x \sim x\\
\arcsin x \sim x& \arctan x \sim x&\\
\ln (1 + x)\sim x& e^x - 1 \sim x& a^x - 1 \sim x \ln a\\
(1 + x)^{\alpha} - 1 \sim \alpha x&&\\
\left(1 + \frac{1}{n}\right)^n = e - \frac{e}{2n} + o(\frac{1}{n})&&
\end{array}
\]
一些证明¶
命题. \(\left( 1 + \frac{1}{n} \right)^n = e - \frac{e}{2n} + o(\frac{1}{n})\)
证明：考虑 \((1 + \frac{1}{n})^n = e^{n \ln(1 + \frac{1}{n})}\) ，对 \(\ln(1 + \frac{1}{n})\) 进行 常用 Taylor 展开 得到
\[ e^{n \ln(1 + \frac{1}{n})} = e^{n \sum\limits_{k = 1}^{+\infty} (-1)^{k-1}\frac{1}{kn^{k}}} = e^{\sum\limits_{k = 1}^{+\infty} (-1)^{k-1} \frac{1}{kn^{k-1}}} \]展开前几项即可得到 \(e^{1 - \frac{1}{2n}} = e(e^{-\frac{1}{2n}}) = e(1 - \frac{1}{2n}) = e - \frac{e}{2n}\)。