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I need to find the limit of the following expression: $$\lim_{x \rightarrow \infty} \left(\left(\frac{x+1}{x-1}\right)^x -e^2\right){x}^2.$$

I got the limit of the first term as $e^2$ so the problem now becomes equivalent to evaluating a $0 \times \infty$ limit. But im having difficulty proceeding after this. I tried applying L'Hospital's Rule but the derivative gets really complicated after a while. I also tried applying series expansions but that too got me nowhere. Could anyone please tell me how to proceed with this problem.

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Hint. Note that for $x>1$, $$\begin{align}\left(\frac{x+1}{x-1}\right)^x&= \exp\left(x\left(\ln\left(1+\frac{1}{x}\right)-\ln\left(1-\frac{1}{x}\right)\right)\right)\\&=\exp\left(2+\frac{2}{3x^2}+o(1/x^2))\right)=e^2\cdot \exp\left(\frac{2}{3x^2}+o(1/x^2))\right)\end{align}$$ where we used the expansion of $\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+o(t^3)$ at $t=0$. Hence $$\lim_{x \rightarrow \infty} \left(\left(\frac{x+1}{x-1}\right)^x -e^2\right){x}^2=e^2\lim_{x \rightarrow \infty} \frac{\exp\left(\frac{2}{3}(1/x^2)+o(1/x^2)\right)-1}{1/x^2}.$$ Can you take it from here?

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  • $\begingroup$ Yes i definitely can take it from here. Thanks a lot! $\endgroup$ – physics123 Mar 13 '18 at 13:38
  • $\begingroup$ @Isham how can we evaluate directly without taylor? $\endgroup$ – physics123 Mar 13 '18 at 13:40
  • $\begingroup$ @physics123 I thought you already had its value as $e^2$ ? $\endgroup$ – Isham Mar 13 '18 at 13:41
  • $\begingroup$ @Isham first term is $e^2$ and first factor tends to $0$. Thats what physics123 found. $\endgroup$ – King Tut Mar 13 '18 at 13:42
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    $\begingroup$ @Isham The result is not $e^2$... $\endgroup$ – Robert Z Mar 13 '18 at 13:44

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