Evaluate $\lim_{x \to \infty} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$ $$\underset{x\to \infty}{\lim} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
My Attempt:
$$L = \underset{t\to 0}{\lim} \frac{\left( \left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2\right)} {t^2}$$
I Now have a $\frac 00$ form that I could use L'Hopital rule with, but I don't want to differentiate the ugly looking function in the numerator. Is there an easier way to solve these kinds of problems? Maybe a taylor series expansion for $(1+t)^{\frac 1t}, t \to 0$ forms would come in handy here and I could just subtract the $e^2$ from the resulting expansion.

$$ \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
Start with
$$y=\left( \frac{x+1}{x-1} \right)^x\implies \log(y)=x\log\left( \frac{x+1}{x-1} \right)=x\log\left(1+ \frac{2}{x-1} \right)$$ Now, by Taylor
$$\log(y)=x\left(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+O\left(\frac{1}{x^7}\right)\right)={2}+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$
$$y=e^{\log(y)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45
   x^4}+O\left(\frac{1}{x^6}\right)$$
$$(y-e^2)x^2=\frac{2 e^2}{3}+\frac{28 e^2}{45 x^2}+O\left(\frac{1}{x^4}\right)$$
