Let L=$\lim_{x\to \infty} \left(\left(\frac{x+1}{x-1}\right)^x-e^2\right)x^2.$Then find the value of $\frac{9L}{2e^2}$.
Following is my approach:
We can see that $\left(\frac{x+1}{x-1}\right)^x$ can be written as $\left(\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}\right)^x \implies \left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)^x$,it is clear that as $x\to \infty$ it is of the form $1^{\infty}$.
So we can use the standard for to solve limits of the type $1^{\infty}$ i.e. if $A=\lim_{x\to a}f(x)^{g(x)}$,where $\lim_{x\to a}f(x)=1$ and $lim_{x\to a}g(x)=\infty$, then $$A=e^{\lim_{x\to a}g(x)[f(x)-1]}$$
Then everything boils down as:
$$L=\lim_{x\to \infty} \left(e^{x\left(\frac{x+1}{x-1}-1\right)}-e^2\right)x^2$$
$$\implies L=\lim_{x\to \infty} \left(e^{x\left(\frac{x+1-x+1}{x-1}\right)}-e^2\right)x^2$$
$$\implies L=\lim_{x\to \infty} \left(e^{\frac{2x}{x-1}}-e^2\right)x^2$$
$$\implies L=\lim_{x\to \infty} \left(e^{\frac{2(x-1+1)}{x-1}}-e^2\right)x^2$$
$$\implies L=\lim_{x\to \infty} \left(e^{2+\frac{2}{x-1}}-e^2\right)x^2$$
$$\implies L=\lim_{x\to \infty} e^2\left(e^{\frac{2}{x-1}}-1\right)x^2$$
$$\implies L=\lim_{x\to \infty} e^2\left(1+\frac{2}{x-1}+\frac{2}{(x-1)^2}....-1\right)x^2 \enspace \enspace \enspace(\because e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}...)$$
$$\implies L=\lim_{x\to \infty} e^2\left(\frac{2x^2}{x-1}+\frac{2x^2}{(x-1)^2}....\right)$$.
Now it can be seen from above that limit is approaching infinity and hence it does not have a finite value.This is where I am stuck as I cannot understand whether my reasoning is valid or not.More over the author has provided a two step solution which went above my head even though I tried hard to understand it.Following is the author's solution:
$$L=\lim_{x\to \infty}\frac{e^2\left(e^{x\ln(\frac{x+1}{x-1})-2}-1\right) \left(x\ln(\frac{x+1}{x-1})-2\right) }{\left(\frac{1}{x^2}\right )\left(x\ln(\frac{x+1}{x-1})-2\right)}$$
$$\implies L=\frac{2}{3}e^2 \implies \frac{9L}{2e^2}=3$$.
I literally did not understand anything what the author did. I would be glad if someone point out the fault in my approach.
The standard approach for handling indeterminates of the form $1^\infty$ is to write
$$e^{\ln \left(f(x)^{g(x)}\right)}=e^{g(x)\ln(f(x))},$$
rather than the expression you used.