Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$

Question

How to calculate the following limit? $$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$

Answer 1

$$\lim_{x\to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}=\lim_{x\to \infty} \Big(1+\frac{2}{x^2-1}\Big)^{\frac{x^2-1}{2}\cdot\frac{2x^2}{x^2-1}}=e^{\lim_{x\to \infty}\frac{2x^2}{x^2-1}}=e^2$$

Answer 2

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.

Answer 3

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 $$ \begin{align} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}&=\Big(\frac{x^2-1+2}{x^2-1}\Big)^{x^2}\\\\ &= \left(1+\frac {2}{x^2-1}\right)^{x^2}\\\\ &=e^{\large x^2\ln \left(1+\frac {2}{x^2-1}\right)}\\\\ &=e^{\large 2+\mathcal{O}(\frac {1}{x^2})}\\\\ &\to e^2 \end{align} $$

Answer 4

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.

Answer 5

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 .. !

Answer 6

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 $t=y/2$.

Answer 7

Using the property that: $e = lim_{x\to\infty} (1 + \frac{1}{x})^{x}$

$lim_{x\to\infty} (\frac{x^{2} + 1}{x^{2} - 1})^{x^{2}}$ = $lim_{x\to\infty} (1 + \frac{2}{x^{2} - 1})^{x^{2}}$ = $e^{lim_{x\to\infty} \frac{2x^{2}}{x^2 - 1}}$

= $e^{2}$