【高等数学笔记-极限(5)】两个重要极限 极限存在准则

第一重要极限

$$\boxed{\underset{x\to 0}{\lim }\frac{\sin x}{x}=1}$$

易得以下极限
$$\boxed{\underset{x\to 0}{\lim }\frac{x}{\sin x}=1}\boxed{\underset{x\to 0}{\lim }\frac{\tan x}{x}=1}$$
例1
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{\tan x}{x}\\ =& \underset{x\to 0}{\lim }\frac{\sin x}{\cos x}\cdot \frac{1}{x}\\ =& \underset{x\to 0}{\lim }\frac{\sin x}{x}\cdot \frac{1}{\cos x}\\ =& 1\times 1=1\end{array}$$
例2
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}\\ =& \underset{x\to 0}{\lim }\frac{(1-\cos x)(1+\cos x)}{x^2\cdot (1+\cos x)}\\ =& \underset{x\to 0}{\lim }\frac{1-{\cos }^{2}x}{x^2\cdot (1+\cos x)}\\ =& \underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{x^2}\cdot \underset{x\to 0}{\lim }\frac{1}{1+\cos x}\\ =& 1^2\times \frac{1}{2}=\frac{1}{2}\end{array}$$
或者使用二倍角公式
$$\cos 2\theta =co{s}^2\theta -{\sin }^2\theta =1-2{\sin }^2\theta$$
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}\\ =& \underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{x}{2}}{x^2}\\ =& \frac{1}{2}\cdot \underset{x\to 0}{\lim }\frac{{\sin }^2(\frac{x}{2})}{(\frac{x}{2}{)}^2}\\ =& \frac{1}{2}\cdot \underset{u\to 0}{\lim }{\left(\frac{\sin u}{u}\right)}^2\\ =& \frac{1}{2}\times 1^2=\frac{1}{2}\end{array}$$
例3
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{\sin \omega x}{x}=\omega \cdot \underset{x\to 0}{\lim }\frac{\sin \omega x}{\omega x}=\omega \cdot \underset{u\to 0}{\lim }\frac{\sin u}{u}=\omega \end{array}$$

例4
$$\underset{x\to 0}{\lim }\frac{\sin 2x}{\sin 5x}=\underset{x\to 0}{\lim }\frac{\frac{\sin 2x}{x}}{\frac{\sin 5x}{x}}=\frac{2}{5}$$

注: $x\to 0⟹x\ne 0$故可以同除以$x$

例4
$$\underset{x\to 0}{\lim }x\cot x=\underset{x\to 0}{\lim }x\cdot \frac{\cos x}{\sin x}=\underset{x\to 0}{\lim }\cos x\cdot \frac{x}{\sin x}=1$$

例5
$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{1-\cos 2x}{x\sin x}\\ =& \underset{x\to 0}{\lim }\frac{1-1+2{\sin }^{2}x}{x\sin x}\\ =& \underset{x\to 0}{\lim }\frac{2\sin x}{x}\\ =& 2\end{array}$$

例6
$$\begin{array}{rl}& \underset{x\to 0}{\lim }{2}^n\sin \frac{x}{2^n}\\ =& \underset{x\to 0}{\lim }{2}^n\cdot \sin \frac{x}{2^n}\cdot \frac{2^n}{x}\cdot \frac{x}{2^n}\\ =& \underset{x\to 0}{\lim }{2}^n\cdot \frac{x}{2^n}\cdot \frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}}\\ =& \underset{x\to 0}{\lim }x\cdot \frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}}\\ =& 0\times \underset{u\to 0}{\lim }\frac{\sin u}{u}=0\times 1=0\end{array}$$

例7
$$\begin{array}{rl}& \underset{n\to \infty }{\lim }{2}^n\sin \frac{x}{2^n}\\ =& \underset{n\to \infty }{\lim }{2}^n\cdot \frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}}\cdot \frac{x}{2^n}\\ =& \underset{n\to \infty }{\lim }x\cdot \frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}}\\ =& x\cdot \underset{u\to 0}{\lim }\frac{\sin u}{u}\\ =& x\times \underset{u\to 0}{\lim }\frac{\sin u}{u}=x\times 1=x\end{array}$$

第二重要极限

$$\boxed{\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e}$$

• 易得以下极限:
  $$\boxed{\underset{x\to 0}{\lim }{\left(1+x\right)}^{\frac{1}{x}}=e}\boxed{\underset{x\to \infty }{\lim }{\left(1-\frac{1}{x}\right)}^x=\frac{1}{e}}$$

$$\begin{gathered} 求\underset{x\to \infty }{\lim }{\left(1-\frac{1}{x}\right)}^x,令u=-x,x\to \infty ,u\to \infty 则有: \\ u\to \end{gathered}$$