例题#

1. 证明 $\sqrt[n]{1+x}-1\sim \dfrac{1}{n}x\quad (x\to 0)$

   证：

   $$\begin{align*} \lim_{x\to 0}\dfrac{\sqrt[n]{1+x}-1}{\frac{1}{n}x}&=\lim_{x\to 0}\dfrac{1+x-1}{\frac{1}{n}\times \left[(1+x)^{\frac{n-1}{n}}+(1+x)^{\frac{n-2}{n}+\cdots + 1}\right]}\\ &=\lim_{x\to 0}\frac{n}{n}=1 \end{align*}$$

2. 证明 $e^x-1\sim x\quad (x\to 0)$

   证：$\displaystyle \lim_{x\to 0}\dfrac{e^x-1}{x}=\lim_{t\to 0}\dfrac{t}{\ln(t+1)}=\lim_{t\to 0}\dfrac{1}{\ln (t+1)^\frac{1}{t}}=\dfrac{1}{\ln e}=1$

3. $\sqrt{x^3+a}-\sqrt{a}\quad (a\gt 0)$ 是 $x$ 的几阶无穷小？

   解：$\displaystyle \lim_{x\to 0}\dfrac{\sqrt{x^3+a}-\sqrt{a}}{x^k}=\lim_{x\to 0}\dfrac{x^3}{x^k(\sqrt{x^3+a}+\sqrt{a})}$

   • 当$k\gt 3$时，极限无穷大
   • 当$k\lt 3$时，极限是0
   • $k=3$时，极限 $\frac{1}{2\sqrt{a}}\ne 0$，3阶无穷小

4. 计算下列极限值

   • $\displaystyle \lim_{x\to 0}\dfrac{\sin 5x}{x}=\lim_{x\to 0}\dfrac{5x}{x}=5$
   • $\displaystyle \lim_{x\to 0}\dfrac{\sin 5x}{\tan 3x}=\lim_{x\to 0}\dfrac{5x}{3x}=\dfrac{5}{3}$
   • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos(1-\cos\frac{x}{2})}{x^4}=\lim_{x\to 0}\dfrac{\frac{1}{2}(1-\cos \frac{x}{2})^2}{x^4}=\lim_{x\to 0}\dfrac{\frac{1}{2}x[\frac{1}{2}(\frac{x}{2})^2]^2}{2}=\dfrac{1}{2^7}$
   • $\displaystyle \lim_{x\to 0}\dfrac{\tan x-\sin x}{x^3}=\lim_{x\to 0}\dfrac{\sin x(\frac{1}{\cos x}-1)}{x^3}=\lim_{x\to 0}\dfrac{x(1-\cos x)}{x^3\cos x}=\lim_{x\to 0}\dfrac{\frac{1}{2}x^3}{x^3\cos x}=\dfrac{1}{2}$
   • $\displaystyle \lim_{x\to 0}\dfrac{\sin 2x-\sin x}{x}=\lim_{x\to 0}\dfrac{\sin x(2\cos x-1)}{x}=1$
   • $\displaystyle \lim_{\alpha \to \beta}\dfrac{e^\alpha - e^\beta}{\alpha - \beta}=\lim_{\alpha \to \beta}\dfrac{e^\beta(e^{\alpha-\beta}-1)}{\alpha - \beta}=e^\beta$