Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
 A: $$\lim_{x\to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}=\lim_{x\to \infty} \Big(1+\frac{2}{x^2-1}\Big)^{\frac{x^2-1}{2}\cdot\frac{2x^2}{x^2-1}}=e^{\lim_{x\to \infty}\frac{2x^2}{x^2-1}}=e^2$$
A: it's the same as that of
$$
\left(\frac{1+1/n}{1-1/n}\right)^{n}
$$
Without the bottom, this gives $e$ as is well known.
$(1-1/n)^{-1} = 1+ 1/n + 1/n^2 +...$
multiplying we get 
$$
(1 + 2/n + O(n^{-2}))^n
$$
If we drop the $O$ term we get $e^2.$ 
it simply remains to show that the final error disappears in the limit. 
A: Recall that, as $u \to 0$, by the Taylor series expansion, we readily have
$$
\begin{align}
e^u& =1+u+\mathcal{O}(u^2)\\
\ln (1+u)&=u+\mathcal{O}(u^2)
\end{align}
$$ giving, as $x \to \infty$,
$$
x^2\ln \left(1+\frac {2}{x^2-1}\right)=x^2 \left(\frac {2}{x^2-1}+\mathcal{O}\left(\frac {1}{x^4}\right)\right)=2+\mathcal{O}\left(\frac {1}{x^2}\right)
$$ and
$$
\begin{align}
\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}&=\Big(\frac{x^2-1+2}{x^2-1}\Big)^{x^2}\\\\
&= \left(1+\frac {2}{x^2-1}\right)^{x^2}\\\\
&=e^{\large x^2\ln \left(1+\frac {2}{x^2-1}\right)}\\\\
&=e^{\large 2+\mathcal{O}(\frac {1}{x^2})}\\\\
&\to e^2
\end{align}
$$
A: You can do it using L'Hopital's rule this way:
$$\begin{align}
\lim\limits_{x \to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} = \lim\limits_{x \to \infty} \exp\ln\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} &= \lim\limits_{x \to \infty} \exp x^2\ln\Big(\frac{x^2+1}{x^2-1}\Big) \\ &= \exp \lim\limits_{x \to \infty} \frac{\ln(x^2+1)-\ln(x^2-1)}{x^{-2}} \\ &= \exp \lim\limits_{x \to \infty} \frac{2x(x^2+1)^{-1}-2x(x^2-1)^{-1}}{-2x^{-3}} \\ &= \exp \lim\limits_{x \to \infty} \frac{x^4(x^2+1)-x^4(x^2-1)}{(x^2+1)(x^2-1)} \\ &= \exp \lim\limits_{x \to \infty} \frac{2x^4}{(x^2+1)(x^2-1)} \\ &= e^2
\end{align}$$
It's permissible to swap the $\exp$ with the $\lim$ between lines one and two because the exponential function is continuous, and L'Hopital's rule is applied between lines two and three.
A: Let's say we know the definition of $e$ as $\displaystyle e=\lim_{x\to \infty }\Big(1+\frac{1}{x}\Big)^x$
\begin{align}
\lim_{x\to \infty }\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}&=\lim_{x\to \infty }\Big(1+\frac{2}{x^2-1}\Big)^{x^2}\\
&=\lim_{y\to \infty }\Big(1+\frac{2}{y}\Big)^{y+1}\\
&=\lim_{y\to \infty }\Big(1+\frac{2}{y}\Big)^{1} \times \lim_{y\to \infty }\Big(1+\frac{2}{y}\Big)^{y}\\
&=\lim_{y\to \infty }\Big(1+\frac{1}{y/2}\Big)^{y}\\
&=\lim_{t\to \infty }\Big(1+\frac{1}{t}\Big)^{2t}\\
&=\Big[\lim_{t\to \infty }\Big(1+\frac{1}{t}\Big)^{t}\Big]^2\\
&=e^2.
\end{align}
Note: $y=x^2$ then later $t=y/2$.
A: Using the property that: $e = lim_{x\to\infty} (1 + \frac{1}{x})^{x}$
$lim_{x\to\infty} (\frac{x^{2} + 1}{x^{2} - 1})^{x^{2}}$ = $lim_{x\to\infty} (1 + \frac{2}{x^{2} - 1})^{x^{2}}$ = $e^{lim_{x\to\infty} \frac{2x^{2}}{x^2 - 1}}$
= $e^{2}$
A: The qty in bracket tends to 1 as x→infinte and power tends to infinity u can easily prove that Lt(x→a){f(x)}^(g(x)) if f(a)→1 and g(a)→∞ then its equal to limit of e^(f(x)-1)(g(x)) as x→a so here it is.. e^(2/(x^2-1))(x^2) limit as x→ ∞ giving e^2 .. !
