问题及解答

证明:

\[\lim_{x\rightarrow 0}x\ln(\sin x)=0.\]

\[\lim_{x\rightarrow\frac{\pi}{2}}\ln(\cos x)\cdot\ln(\sin x)=0.\]

\[
\lim_{x\rightarrow 0}x\ln(\sin x)=\lim_{x\rightarrow 0}\frac{\ln(\sin x)}{\frac{1}{x}}
\]

应用洛必达法则即可求出.

(2)

\[
\begin{split}
\lim_{x\rightarrow\frac{\pi}{2}}\ln(\cos x)\cdot\ln(\sin x)&=\lim_{x\rightarrow\frac{\pi}{2}}\ln(\cos x)\cdot\ln(1+(\sin x-1))\\
&=\lim_{x\rightarrow\frac{\pi}{2}}\ln(\cos x)\cdot(\sin x-1)\\
&=\lim_{x\rightarrow\frac{\pi}{2}}\frac{\ln(\cos x)}{\frac{1}{\sin x-1}}\\
&=\lim_{x\rightarrow\frac{\pi}{2}}\frac{\frac{-\sin x}{\cos x}}{-\frac{1}{(\sin x-1)^2}\cdot\cos x}\\
&=\lim_{x\rightarrow\frac{\pi}{2}}\frac{(\sin x-1)^2}{\cos^2 x}
\end{split}
\]

而

\[
\lim_{x\rightarrow\frac{\pi}{2}}\frac{\sin x-1}{\cos x}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos x}{-\sin x}=0,
\]

故原极限为 0.  Q.E.D.