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

If you switch variables to $y = x + 1$ you are trying to get this limit:
$$\lim_{y \rightarrow \infty} \bigg(1 + {2 \over y}\bigg)^{\sqrt{y^2 + 2y}}$$
$$= \lim_{y \rightarrow \infty} \bigg(1 + {2 \over y}\bigg)^{y\sqrt{1 + {2 \over y}}}$$
Taking logs finishes it pretty quick...
$$\lim_{y \rightarrow \infty} y\sqrt{1 + {2 \over y}}\ln(1 + {2 \over y}) $$
$$= \lim_{y \rightarrow \infty} y\ln(1 + {2 \over y})$$
$$= \lim_{y \rightarrow \infty} {\ln(1 + {2 \over y}) \over {1 \over y}}$$
$$= \lim_{z \rightarrow  0} {\ln(1 + {2 z}) \over {z}}$$
Now apply L'hopital. You could also have applied l'Hopital after the previous step.
