$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ 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.
 A: rewrite your term $y=\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}$ in the form $e^{\frac{\ln\left(\frac{(x+1)}{x-1}\right)}{\frac{1}{\sqrt{x^2-1}}}}$
 and use L'Hospital.
A: 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 $a,b>0$.
A: $$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}=\left(1+\frac2{x-1}\right)^{\sqrt{x^2-1}}=\exp\left(\sqrt{x^2-1}\;\log\left(1+\frac2{x-1}\right)\right)$$
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:
$$\lim_{x\to \infty}\frac{\log\left(1+\frac2{x-1}\right)}{\frac1{\sqrt{x^2-1}}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac{x-1}{x+1}\left(-\frac2{(x-1)^2}\right)}{-\frac x{(x^2-1)\sqrt{x^2-1}}}=$$
$${}$$
$$=\lim_{x\to\infty}\frac{2\sqrt{x^2-1}}{x}=\lim_{x\to\infty}\frac{2\sqrt{1-\frac1{x^2}}}{1}=2\;\;\implies$$
$$\lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}=\exp(2)=e^2$$
A: 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.
A: Note that:
$$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = \left(1 + \frac{2}{x-1}\right)^{\sqrt{x^2-1}} = e^{\sqrt{x^2-1} \log \left(1 + \frac{2}{x-1}\right)}$$
Now we work with the exponent:
$$\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \log \left(1 + \frac{2}{x-1}\right)
\overset{(1)}{=} 
\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \frac{2}{x-1} 
\overset{(2)}{=} 
2$$
Therefore:
$$\lim_{x\rightarrow \infty} \left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = e^2$$

Observations.
$(1)$ Note that $\log(1+z) \sim z$ when $z$ goes to zero. Here "$z = \frac{2}{x-1}$".
$(2)$ Clearly $\lim_{x\rightarrow \infty} \frac{\sqrt{x^2-1}}{x-1}=
\lim_{x\rightarrow \infty}\sqrt{\frac{(x-1)(x+1)}{(x-1)^2}}=
\lim_{x\rightarrow \infty} \sqrt{\frac{x+1}{x-1}}=
\lim_{x\rightarrow \infty} \sqrt{1+\frac{2}{x-1}}
=1$.
