<s>[INST] $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ 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.
 [/INST] 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 $\lim_{x\rightarrow \infty} \frac{\sqrt{x^2-1}}{x-1}=
\lim_{x\rightarrow \infty}\sqrt{\frac{(x-1)(x+1)}{(x-1)^2}}=
\lim_{x\rightarrow \infty} \sqrt{\frac{x+1}{x-1}}=
\lim_{x\rightarrow \infty} \sqrt{1+\frac{2}{x-1}}
=1$.
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