How to evaluate $\lim_{x \rightarrow \infty} \left(\left(\frac{x+1}{x-1}\right)^x -e^2\right){x}^2$

Let $t=\frac1x$.

We have that

$$\frac{2t}{1-t}=2t+2t^2+2t^3+O(t^4)$$

then

$$\left(\frac{1+t}{1-t}\right)^{1/t}=e^{\frac1t\log\left(1+\frac{2t}{1-t}\right)}=e^{\frac1t\left(2t+\frac23t^3+O(t^4)\right)}=e^2\left(1+\frac23t^2+O(t^3)\right)$$

and therefore

$$\frac{\left(\frac{1+t}{1-t}\right)^{1/t}-e^2}{t^2}=\frac{e^2\left(1+\frac23t^2+O(t^3)\right)-e^2}{t^2}$$

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?

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