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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?

joseabp91
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    Something isn't quite right here: The limit, as given, does not exist, because $1-x^2$ is negative for $x\gt1$, which means the expression is undefined for most values of $x$. And even if you restrict it, say to integer values of $x$, it'll be positive for even integer values and negative for odd integer values, so its limit, if one did exist, could only be $0$. I think you need to have $x^2-1$ in the denominator. – Barry Cipra Jan 26 '21 at 16:33
  • See https://math.stackexchange.com/questions/580665/evaluate-lim-limits-n-to-infty-left-fracn-1-2n2-rightn/580667#580667 and https://math.stackexchange.com/questions/3071229/simple-way-to-compute-lim-x-to-0-left-frac1x12x-right1-x-witho/3071235#3071235 – lab bhattacharjee Jan 26 '21 at 16:40
  • Probably it is $$\underset{x\to \infty }{\text{lim}}\left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$ – Raffaele Jan 26 '21 at 16:41
  • Furthermore $x^{x^2}\ne (x^x)^2$ – Raffaele Jan 26 '21 at 16:43
  • @BarryCipra If you are right then the limit is trivially $1$. The limit I wrote is $e^2$ – Raffaele Jan 26 '21 at 16:53
  • why is the limit you wrote $e^2$? – joseabp91 Jan 26 '21 at 16:59
  • @Raffaele, oh, you are right. I'll delete that unfortunate comment! – Barry Cipra Jan 26 '21 at 17:04
  • If I replace $1-x^2$ by $x^2-1$ from the denominator, how can I check that I get $e^2$? – joseabp91 Jan 26 '21 at 17:14

2 Answers2

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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?

Mark Viola
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Hint: divide both denominator and numerator by $x^2$ and rewrite the quotient as a product where each factor tends to $e$.

DeepSea
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