求极限 $\lim\limits_{h\rightarrow 0}\displaystyle\frac{1}{h^2}\int_0^h \Bigl(\frac{1}{\theta}-\cot\theta\Bigr)\mathrm{d}\theta$.
求极限
\[
\lim_{h\rightarrow 0}\frac{1}{h^2}\int_0^h \Bigl(\frac{1}{\theta}-\cot\theta\Bigr)\mathrm{d}\theta
\]
Answer
问题及解答
1
\[
\begin{split}
\lim_{h\rightarrow 0}\frac{1}{h^2}\int_0^h \Bigl(\frac{1}{\theta}-\cot\theta\Bigr)\mathrm{d}\theta&\xlongequal[\text{洛}]{\frac{0}{0}}\lim_{h\rightarrow 0}\frac{\frac{1}{h}-\cot h}{2h}=\lim_{h\rightarrow 0}\frac{\frac{1}{h}-\frac{1}{\tan h}}{2h}=\lim_{h\rightarrow 0}\frac{\tan h-h}{2h^2\tan h}\\
&=\lim_{h\rightarrow 0}\frac{\tan h-h}{2h^3}\xlongequal[\text{洛}]{\frac{0}{0}}\lim_{h\rightarrow 0}\frac{\sec^2 h-1}{6h^2}=\lim_{h\rightarrow 0}\frac{\tan^2 h}{6h^2}\\
&=\lim_{h\rightarrow 0}\frac{h^2}{6h^2}=\frac{1}{6}.
\end{split}
\]