P. 44--45
4. 求下列极限
(4) $\lim\limits_{x\rightarrow\frac{\pi}{2}}\dfrac{\cos x}{2x-\pi}$
(5) $\lim\limits_{x\rightarrow\pi}\dfrac{\sin x}{x-\pi}$
(7) $\lim\limits_{x\rightarrow 0^-}\dfrac{x}{\sqrt{1-\cos x}}$
5. 求下列极限
(4) $\lim\limits_{x\rightarrow 1}(3-2x)^{\frac{3}{x-1}}$
(6) $\lim\limits_{x\rightarrow\infty}\Bigl(\frac{x+2}{x-2}\Bigr)^x$
(8) $\lim\limits_{x\rightarrow\infty}\biggl(\frac{n^2+3}{n^2+1}\biggr)^{n^2}$
1
4. 求下列极限
(4) $\lim\limits_{x\rightarrow\frac{\pi}{2}}\dfrac{\cos x}{2x-\pi}$
解:
\[
\text{原式}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{\sin(\frac{\pi}{2}-x)}{2(x-\frac{\pi}{2})}=-\frac{1}{2}
\]
(5) $\lim\limits_{x\rightarrow\pi}\dfrac{\sin x}{x-\pi}$
解:
\[
\text{原式}=\lim_{x\rightarrow\pi}\frac{\sin(\pi-x)}{x-\pi}=-1
\]
(7) $\lim\limits_{x\rightarrow 0^-}\dfrac{x}{\sqrt{1-\cos x}}$
解:
\[
\text{原式}=\lim_{x\rightarrow 0^-}\frac{x}{\sqrt{2\sin^2\frac{x}{2}}}=\lim_{x\rightarrow 0^-}\frac{2\cdot\frac{x}{2}}{\sqrt{2}\cdot\sin\frac{|x|}{2}}=\sqrt{2}\cdot\lim_{x\rightarrow 0^-}\frac{-\frac{|x|}{2}}{\sin\frac{|x|}{2}}=-\sqrt{2}
\]
2
5. 求下列极限
(4) $\lim\limits_{x\rightarrow 1}(3-2x)^{\frac{3}{x-1}}$
解:
\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow 1}\big(1+2(1-x)\bigr)^{\frac{3}{x-1}}\\
&=\lim_{x\rightarrow 1}\biggl[\Big(1+2(1-x)\Bigr)^{\dfrac{1}{2(1-x)}}\biggr]^{-6}\\
&=\biggl[\lim_{x\rightarrow 1}\Big(1+2(1-x)\Bigr)^{\dfrac{1}{2(1-x)}}\biggr]^{-6}\\
&=e^{-6}
\end{split}
\]
(6) $\lim\limits_{x\rightarrow\infty}\Bigl(\frac{x+2}{x-2}\Bigr)^x$
解:
\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow\infty}\Bigl(1+\frac{4}{x-2}\Bigr)^x\\
&=\lim_{x\rightarrow\infty}\Biggl\{\biggl[\Bigl(1+\frac{4}{x-2}\Bigr)^{\frac{x-2}{4}}\biggr]^4\cdot(1+\frac{4}{x-2})^2\Biggr\}\\
&=e^4\cdot 1\\
&=e^4
\end{split}
\]
(8) $\lim\limits_{x\rightarrow\infty}\biggl(\frac{n^2+3}{n^2+1}\biggr)^{n^2}$
解:
\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow\infty}\biggl(1+\frac{2}{n^2+1}\biggr)^{n^2}\\
&=\lim_{x\rightarrow\infty}\Biggl\{\biggl[\Bigl(1+\frac{2}{n^2+1}\Bigr)^{\frac{n^2+1}{2}}\biggr]^{2}\cdot(1+\frac{2}{n^2+1})^{-1}\Biggr\}\\
&=e^2\cdot 1\\
&=e^2
\end{split}
\]