MA.3.4 未定式的极限

$\underset{x\to 0}{lim}{\displaystyle \frac{x\cos x-\sin x}{x^3}}$

Solution:
$\stackrel{\frac{0}{0}}{=}\underset{x\to 0}{lim}{\displaystyle \frac{-x\sin x}{3x^2}}=-{\displaystyle \frac{1}{3}}$

$\underset{x\to 0}{lim}{\displaystyle \frac{e^x+x^{-x}-2}{{\sin }^{2}x}}$

Solution:
$=\underset{x\to 0}{lim}{\displaystyle \frac{e^x+e^{-x}-2}{x^2}}\stackrel{\frac{0}{0}}{=}\underset{x\to 0}{lim}{\displaystyle \frac{e^x-e^{-x}}{2x}}\stackrel{\frac{0}{0}}{=}\underset{x\to 0}{lim}{\displaystyle \frac{e^x+e^{-x}}{2}}=1$

$\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{x^{\mu }}{a^x}}=0(a>1)$

Proof:
$1.\mu =0:\mathrm{显}\mathrm{然}\mathrm{为}0$
$2.\mu \ne 0$
$\sim {\displaystyle \frac{\mu x^{\mu -1}}{a{x}^{\mu -1}}}\sim \cdots \sim {\displaystyle \frac{\mu (\mu -1)(\mu -2)\dots (\mu -n+1)a^{\mu -n}}{a^x{\ln }^{n}a}}=0$
P.S. $n(\mu )\mathrm{此}\mathrm{时}\mathrm{是}\mathrm{常}\mathrm{数}$

$$a^x>x^{\alpha }>\ln x\dots (a>1,\alpha >0)$$

$\underset{x\to +\mathrm{\infty }}{lim}x({\displaystyle \frac{\pi }{2}}-\arcsin x)$

Solution:
$\begin{array}{rl}& =\underset{x\to +\mathrm{\infty }}{lim}x(\frac{\pi }{2}-\arctan x)\\ & =\underset{x\to +\mathrm{\infty }}{lim}\frac{\frac{\pi }{2}-\arctan x}{x^{-1}}\\ & \stackrel{\frac{0}{0}}{=}\underset{x\to +\mathrm{\infty }}{lim}\frac{\frac{1}{1+x^2}}{\frac{1}{x^2}}\\ & =\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2}{1+x^2}\\ & =1\end{array}$

$\underset{x\to 0^{+}}{lim}x\ln x$

Solution:
$\begin{array}{rl}& =\underset{x\to 0^{+}}{lim}\frac{\ln x}{\frac{1}{x}}\\ & \stackrel{\sim }{=}\underset{x\to 0^{+}}{lim}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\\ & \stackrel{\frac{\mathrm{\infty }}{\mathrm{\infty }}}{=}\underset{x\to 0^{+}}{lim}(-x)\\ & =0\end{array}$

$\underset{x\to 0^{+}}{lim}{x}^{\alpha }\ln x=0$

$\underset{x\to 1}{lim}(\frac{1}{\ln x}+\frac{1}{1-x})$
$\begin{array}{rl}& =\underset{x\to 1}{lim}\frac{1-x+\ln x}{\ln x(1-x)}\\ & =\underset{x\to 1}{lim}\frac{1-x+\ln x}{(1-x{)}^2}\\ & ={\displaystyle \frac{1}{2}}& \cdots \ln x=\ln (1+x-1)\sim x-1\end{array}$

$\underset{x\to 0}{lim}(\frac{1}{{\sin }^{2}x}-\frac{1}{x^2})$

Solution:
$\begin{array}{rl}& =\underset{x\to 0}{lim}\frac{x^2-{\sin }^{2}x}{{\sin }^{2}x\cdot x^2}\\ & =\underset{x\to 0}{lim}\frac{x^2-{\sin }^{2}x}{x^4}\\ & =\underset{x\to 0}{lim}\frac{(x+\sin x)(x-\sin x)}{x\cdot x^3}\\ & =\underset{x\to 0}{lim}\frac{x+\sin x}{x}\cdot \underset{x\to 0}{lim}\frac{x-\sin x}{x^3}\\ & =2\cdot \underset{x\to 0}{lim}\frac{1-\cos x}{3x^2}\\ & =\frac{1}{3}\end{array}$

$\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}$

Solution:
$\begin{array}{rl}& =e^{\underset{x\to 0}{lim}((\sin x/x-1)\cdot 1/x^2)}\\ & =e^{\underset{x\to 0}{lim}(\sin x-x)/x^3}\\ & =e^{-\frac{1}{6}}\end{array}$

$\begin{array}{rl}& \underset{x\to 0^{+}}{lim}{x}^x\\ & =\underset{x\to 0^{+}}{lim}{e}^{x\ln x}\\ & =e^{\underset{x\to 0^{+}}{lim}(x\ln x)}\\ & =e^0\\ & =1\end{array}$

$\begin{array}{rl}& \underset{x\to +\mathrm{\infty }}{lim}{x}^{\frac{1}{\sqrt{x}}}\\ & y=y(x):=x^{\frac{1}{\sqrt{x}}}\\ & \Rightarrow \ln y=\frac{\ln x}{\sqrt{x}}\\ & \because \underset{x\to +\mathrm{\infty }}{lim}{x}^{\frac{1}{2}}\sqrt{x}=0L^{\prime }Hospital\frac{\mathrm{\infty }}{\mathrm{\infty }}\\ & \therefore \underset{x\to +\mathrm{\infty }}{lim}{x}^{\frac{1}{\sqrt{x}}}=\underset{x\to +\mathrm{\infty }}{lim}{e}^{\ln y}=e^{\underset{x\to +\mathrm{\infty }}{lim}(\ln y)}=e^0=1\end{array}$

#Heine归并定理

$\underset{n\to +\mathrm{\infty }}{lim}{n}^{\sin \frac{1}{n}}$

先求：
$\begin{array}{rl}& \underset{x\to +\mathrm{\infty }}{lim}{x}^{\sin \left(\frac{1}{x}\right)}\\ & =e^{\underset{x\to +\mathrm{\infty }}{lim}\sin \left(\frac{1}{x}\right)\ln x}\\ & =e^{\underset{x\to +\mathrm{\infty }}{lim}\frac{\ln x}{x}}\\ & =e^0\\ & =1\end{array}$

#Heine归并定理

$\begin{array}{rl}& \underset{x\to 0}{lim}(1-\cos x{)}^{\tan x}\\ & =\underset{x\to 0}{lim}{e}^{\tan x\ln (1-\cos x)}\\ & =e^{\underset{x\to 0}{lim}\frac{\ln (1-\cos x)}{\frac{1}{x}}}\\ & =e^{\underset{x\to 0}{lim}\frac{\frac{\sin x}{1-\cos x}}{-\frac{1}{x^2}}}\\ & =e^{\underset{x\to 0}{lim}\frac{x^2\sin x}{-(1-\cos x)}}\\ & =e^{\underset{x\to 0}{lim}-\frac{x^3}{\frac{1}{2}{x}^2}}\\ & =e^0\\ & =1\end{array}$

$★\underset{x\to 0}{lim}\frac{x^2\sin \left(\frac{1}{x}\right)}{\sin x}$

Solution 1:
$\begin{array}{rlr}& =\underset{x\to 0}{lim}\frac{x^2\sin \left(\frac{1}{x}\right)}{x}& \cdots \sin x\sim x\\ & =\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)& \mathrm{有}\mathrm{界}\mathrm{量}\times \mathrm{无}\mathrm{穷}\mathrm{小}\\ & =0\end{array}$

Solution 2:(WRONG)

$\begin{array}{rl}& \underset{x\to 0}{lim}\frac{x^2\sin (\frac{1}{x})}{\sin x}\\ & \stackrel{\frac{0}{0}}{=}\underset{x\to 0}{lim}\frac{2x\sin (\frac{1}{x})-\stackrel{\mathrm{震}\mathrm{荡}}{\cos (\frac{1}{x})}}{\cos x}& \cdots \mathrm{整}\mathrm{个}\mathrm{分}\mathrm{式}\mathrm{无}\mathrm{极}\mathrm{限}\\ & \Rightarrow \mathrm{商}\mathrm{无}\mathrm{极}\mathrm{限}\Rightarrow \mathrm{函}\mathrm{数}\mathrm{无}\mathrm{极}\mathrm{限}\end{array}$

$★f\in D[a,+\mathrm{\infty })\text{且}\underset{x\to +\mathrm{\infty }}{lim}f(x)\text{存在}$
$\text{是否有}\underset{x\to +\mathrm{\infty }}{lim}{f}^{\prime }(x)=0$

Proof:(WRONG)
$\mathrm{构}\mathrm{造}{\displaystyle \frac{f(x)}{x}}$
$(\frac{1}{x}\mathrm{无}\mathrm{穷}\mathrm{小}\times \mathrm{有}\mathrm{界}\mathrm{量})=0=\underset{x\to +\mathrm{\infty }}{lim}\frac{f(x)}{x}\text{⧸}\stackrel{\frac{\mathrm{\infty }}{\mathrm{\infty }}}{=}\underset{x\to +\mathrm{\infty }}{lim}\frac{f^{\prime }(x)}{1}$
$\mathrm{函}\mathrm{数}\mathrm{商}\nRightarrow \mathrm{导}\mathrm{数}\mathrm{商}$
$f^{\prime }(x)$不一定有极限！

反例：构造震荡
$f(x)=\frac{1}{x}\cdot \sin (x^2)$
$\underset{x\to +\mathrm{\infty }}{lim}f(x)=0\dots 0\cdot \mathrm{有}\mathrm{界}\mathrm{量}$
$f^{\prime }(x)=\sin (x^2)+cos(x^2)\cdot 2x\cdot {\displaystyle \frac{1}{x}}$$=\sin (x^2)+2\cos (x^2)$
震荡，无极限