问题及解答

证明:

$$\underset{x\to 0}{lim}x\ln (\sin x)=0.$$

$$\underset{x\to \frac{\pi }{2}}{lim}\ln (\cos x)\cdot \ln (\sin x)=0.$$

$$\underset{x\to 0}{lim}x\ln (\sin x)=\underset{x\to 0}{lim}\frac{\ln (\sin x)}{\frac{1}{x}}$$

应用洛必达法则即可求出.

(2)

$$\begin{array}{rl}\underset{x\to \frac{\pi }{2}}{lim}\ln (\cos x)\cdot \ln (\sin x)& =\underset{x\to \frac{\pi }{2}}{lim}\ln (\cos x)\cdot \ln (1+(\sin x-1))\\ & =\underset{x\to \frac{\pi }{2}}{lim}\ln (\cos x)\cdot (\sin x-1)\\ & =\underset{x\to \frac{\pi }{2}}{lim}\frac{\ln (\cos x)}{\frac{1}{\sin x-1}}\\ & =\underset{x\to \frac{\pi }{2}}{lim}\frac{\frac{-\sin x}{\cos x}}{-\frac{1}{(\sin x-1{)}^2}\cdot \cos x}\\ & =\underset{x\to \frac{\pi }{2}}{lim}\frac{(\sin x-1{)}^2}{{\cos }^{2}x}\end{array}$$

而

$$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sin x-1}{\cos x}=\underset{x\to \frac{\pi }{2}}{lim}\frac{\cos x}{-\sin x}=0,$$

故原极限为 0.  Q.E.D.