计算以下极限:
(1) $\lim _{x \rightarrow \infty} \sqrt{x^2+1}-\sqrt{x^2-1}$;
(2) $\lim _{x \rightarrow \infty}\left(\frac{3 x+2}{3 x+1}\right)^{x+5}$
【答案】 $$
\begin{aligned}
& \lim _{x \rightarrow \infty} \sqrt{x^2+1}-\sqrt{x^2-1} \\
& =\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right)\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}{\sqrt{x^2+1}+\sqrt{x^2-1}} \\
& =\lim _{x \rightarrow \infty} \frac{x^2+1-\left(x^2-1\right)}{\sqrt{x^2+1}+\sqrt{x^2-1}} \\
& =\lim _{x \rightarrow \infty} \frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}}=0
\end{aligned}
$$
(2)
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{3 x+2}{3 x+1}\right)^{x+5} \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{1}{3 x+1}\right)^{x+5} \\
& =\lim _{x \rightarrow \infty}\left(1+\frac{1}{3 x+1}\right)^{(3 x+1) \frac{x+5}{3 x+1}} \\
& =e^{\frac{1}{3}}
\end{aligned}
$$


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