$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule. 
I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having trouble differentiating $f(x) = (\frac{x+1}{x-1})^{\sqrt{x^2-1}}$. I've used Mathematica, but I won't understand it unless I see the step-by-step process.

Use that if $\displaystyle\lim_{x\to+\infty}f(x)=L$ then $\displaystyle\lim_{x\to+\infty}\ln(f(x))=\ln(L)$, for $f(x),L>0$.
And that $\ln(a^b)=b\ln(a)$ for $a,b>0$.
