<s>[INST] Limit to $e^2$. I have to show
$$
\lim_{x \to \infty} {\left(\frac{x^2 + 1}{1 - x^2}\right)}^{x^2} = e^2,
$$
but I don't get the trick to see it, I suppose I can use something like
$$
{\left(\lim_{x \to \infty} {\left(1 + \frac1x\right)}^x\right)}^2 = e^2,
$$
but I do not see how I can apply to the problem. On the other hand,
$$
\frac{x^2 + 1}{1 - x^2} = 1 + \frac{2 x^2}{1 - x^2}.
$$
Any hint?
 [/INST] If the limit of interest is $\lim_{x\to\infty}\left(\frac{x^2+1}{x^2-1}\right)^{x^2}$, then we can write
$$\begin{align}
\left(\frac{x^2+1}{x^2-1}\right)^{x^2}&=\left(\frac{x^2-1+2}{x^2-1}\right)^{x^2}\\\\
&=\left(\left(1+\frac{2}{x^2-1}\right)^{x^2-1}\right)^{x^2/(x^2-1)}
\end{align}$$
Now, use $\lim_{y\to\infty}\left(1+\frac{t}{y}\right)^y=e^t$ with $y=x^2-1$.  Can you wrap this up now?
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