I want 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 I'm 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?


4 Answers 4


Let $t=\frac1x$.

We have that




and therefore


  • $\begingroup$ can you add some detail how are you getting $2/3 t^3$ in second line? $\endgroup$
    – user69608
    Jul 31, 2020 at 6:54
  • 1
    $\begingroup$ We need to use third order expansion for $\log(1+u)$ with $u=2t+2t^2+2t^3+o(t^4)$. $\endgroup$
    – user
    Jul 31, 2020 at 7:06
  • $\begingroup$ Try by yourself, I’ll add these details later if needed! $\endgroup$
    – user
    Jul 31, 2020 at 7:15
  • $\begingroup$ in the last line it should be e^(2{something}) and not e^2 * (something) $\endgroup$ Jul 31, 2020 at 7:16
  • $\begingroup$ I think it’s fine that way, look at the previous steps. $\endgroup$
    – user
    Jul 31, 2020 at 7:18

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?

  • $\begingroup$ Yes i definitely can take it from here. Thanks a lot! $\endgroup$
    – rohit_r
    Mar 13, 2018 at 13:38
  • $\begingroup$ @Isham how can we evaluate directly without taylor? $\endgroup$
    – rohit_r
    Mar 13, 2018 at 13:40
  • $\begingroup$ @physics123 I thought you already had its value as $e^2$ ? $\endgroup$ Mar 13, 2018 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, 2018 at 13:42
  • 1
    $\begingroup$ @Isham The result is not $e^2$... $\endgroup$
    – Robert Z
    Mar 13, 2018 at 13:44

Let $t=1/x$, then the limit becomes $$\begin{aligned}L &=\displaystyle{\lim_ {t\to 0^{+}}}\frac{\left(\frac{1+t}{1-t}\right)^{\frac{1}{t}}-e^2}{t^2} \\ &=\lim_{t\rightarrow0}\frac{e^{\frac{1}{t}\ln\frac{1+t}{1-t}}-e^2}{t^2}\\ &=\lim_{t\rightarrow0}\frac{\left(1+\frac{2t}{1-t}\right)^{\frac{1-t} {2t}\cdot\frac{2t}{(1-t)t}}\left(\frac{1}{t}\ln\frac{1+t}{1-t}\right)'}{2t} \\ &=e^2\lim_{t\rightarrow0}\frac{\left(\frac{1}{t}\ln\frac{1+t}{1-t}\right)'}{2t} \\ &=e^2\lim_{t\rightarrow0}\frac{-\ln\frac{1+t}{1-t}+\frac{2t}{1-t^2}}{2t^3} \\ &=e^2\lim_{t\rightarrow0}\frac{\left(-\ln\frac{1+t}{1-t}+\frac{2t}{1-t^2}\right)'}{6t^2} \\ &=e^2\lim_{t\rightarrow0}\frac{\frac{4t^2}{(1-t^2)^2}}{6t^2} \\ &=\boxed{\frac{2}{3}e^2} \end{aligned}$$

  • $\begingroup$ Step 2 to 3: is it 0/0? $\endgroup$
    – C.F.G
    Jul 31, 2020 at 6:39
  • $\begingroup$ Yes, of course. I added it. $\endgroup$ Jul 31, 2020 at 6:40
  • $\begingroup$ $\frac{e^{\frac{0}{0}}-e^2}{0^2}=0/0?$ $\endgroup$
    – C.F.G
    Jul 31, 2020 at 6:43
  • $\begingroup$ @C.F.G See please better my post. I showed it. See the following expression $\left(1+\frac{2t}{1-t}\right)^{\frac{1-t}{2t}\cdot\frac{2t}{(1-t)t}}$. It closed to $e^2$. $\endgroup$ Jul 31, 2020 at 6:46

As an alternative we have



with the standard limit

$$\frac{e^{\frac1t\log\left(\frac{1+t}{1-t}\right)-2}-1}{\frac1t\log\left(\frac{1+t}{1-t}\right)-2} \to 1$$

and by $\log(1+u)=u-\frac12u^2+\frac13u^3+o(u^3)$

$$\frac{\frac1t\log\left(\frac{1+t}{1-t}\right)-2}{t^2}=\frac{\log(1+t)-\log(1-t)-2t}{t^3} \to\frac23$$


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