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?


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?

| cite | improve this answer | |
  • $\begingroup$ Yes i definitely can take it from here. Thanks a lot! $\endgroup$ – gateway2745 Mar 13 '18 at 13:38
  • $\begingroup$ @Isham how can we evaluate directly without taylor? $\endgroup$ – gateway2745 Mar 13 '18 at 13:40
  • $\begingroup$ @physics123 I thought you already had its value as $e^2$ ? $\endgroup$ – Aryadeva 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
  • 1
    $\begingroup$ @Isham The result is not $e^2$... $\endgroup$ – Robert Z Mar 13 '18 at 13:44

Let $t=\frac1x$.

We have that




and therefore


| cite | improve this answer | |
  • $\begingroup$ can you add some detail how are you getting $2/3 t^3$ in second line? $\endgroup$ – user69608 Jul 31 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 at 7:06
  • $\begingroup$ Try by yourself, I’ll add these details later if needed! $\endgroup$ – user Jul 31 at 7:15
  • $\begingroup$ in the last line it should be e^(2{something}) and not e^2 * (something) $\endgroup$ – Anindya Prithvi Jul 31 at 7:16
  • $\begingroup$ I think it’s fine that way, look at the previous steps. $\endgroup$ – user Jul 31 at 7:18

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

| cite | improve this answer | |
  • $\begingroup$ Step 2 to 3: is it 0/0? $\endgroup$ – C.F.G Jul 31 at 6:39
  • $\begingroup$ Yes, of course. I added it. $\endgroup$ – Michael Rozenberg Jul 31 at 6:40
  • $\begingroup$ $\frac{e^{\frac{0}{0}}-e^2}{0^2}=0/0?$ $\endgroup$ – C.F.G Jul 31 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 $e^2$. $\endgroup$ – Michael Rozenberg Jul 31 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 $\log(1+u)=u-\frac12u^2+\frac13u^3+o(u^3)$

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.