问： 求数列极限 $\displaystyle \lim \sqrt{n^3}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$
解：
习题2.1(14-11),
注意到
$$
\begin{aligned}
\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}
=&\frac{(\sqrt{n+2}+\sqrt{n})^2-(2\sqrt{n+1})^2}{\sqrt{n+2}+\sqrt{n}+2\sqrt{n+1}}\\
=&\frac{2(\sqrt{n(n+2)}-(n+1))}{\sqrt{n+2}+\sqrt{n}+2\sqrt{n+1}}\\
=&\frac{2}{\sqrt{n+2}+\sqrt{n}+2\sqrt{n+1}} \frac{(\sqrt{n(n+2)})^2-(n+1)^2}{\sqrt{n(n+2)}+(n+1)}\\
=&\frac{2}{\sqrt{n+2}+\sqrt{n}+2\sqrt{n+1}} \frac{-1}{\sqrt{n(n+2)}+(n+1)}\\
\end{aligned}
$$
从而有
$$
\begin{aligned}
=&\lim \frac{-2 \sqrt{n}}{\sqrt{n+2}+\sqrt{n}+2\sqrt{n+1}} \frac{n}{\sqrt{n(n+2)}+(n+1)}\\
=&\lim \frac{-2}{\sqrt{1+\frac2n}+1+2\sqrt{1+\frac1n}} \frac{1}{\sqrt{1+\frac2n}+(1+\frac1n)}\\
=&\frac{-1}4
\end{aligned}
$$