$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ 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.

$$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}=\left(1+\frac2{x-1}\right)^{\sqrt{x^2-1}}=\exp\left(\sqrt{x^2-1}\;\log\left(1+\frac2{x-1}\right)\right)$$
Try now l'Hospital with
$$\lim_{x\to \infty}\frac{\log\left(1+\frac2{x-1}\right)}{\frac1{\sqrt{x^2-1}}}\;,\;\;\text{and then use continuity of the exponential function}$$
Caution: it looks like l'H isn't going to work, but after you do some algebraic order in the resulting slightly messy expression it certainly works. 
Added on request:
$$\lim_{x\to \infty}\frac{\log\left(1+\frac2{x-1}\right)}{\frac1{\sqrt{x^2-1}}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac{x-1}{x+1}\left(-\frac2{(x-1)^2}\right)}{-\frac x{(x^2-1)\sqrt{x^2-1}}}=$$
$${}$$
$$=\lim_{x\to\infty}\frac{2\sqrt{x^2-1}}{x}=\lim_{x\to\infty}\frac{2\sqrt{1-\frac1{x^2}}}{1}=2\;\;\implies$$
$$\lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}=\exp(2)=e^2$$
