Evaluating $\lim_{x\to \infty}(\frac{x}{x-1})^x$ I am going over a solution given to solving the follow limit,
$$\lim_{x\to \infty}(\frac{x}{x-1})^x$$
The solution continues as follows,
Consider raising the function to $e^{ln\cdots}$
We can find the limit as follows,
$$\lim_{x\to \infty} x \ln(\frac{x}{x-1}) = \lim_{x\to \infty} \frac{\ln(\frac{x}{x-1})}{\frac{1}{x}}$$
The solution argues this is just $\frac{0}{0}$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $e$. 
However, I don't understand how that expression evaluates to $\frac{0}{0}$, in fact it seems to express
$$\frac{\ln(\frac{\infty}{\infty})}{0}$$
I assume the argument is that $\frac{\infty}{\infty}$ equals 1, and $\ln(1) = 0$, so we have $\frac{0}{0}$. But I thought we cannot evaluate $\frac{\infty}{\infty}$? 

L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
\left(\frac{x }{x-1}\right)^x
=
\left(\frac{y+1}{y}\right)^{y+1}
=
\left(1 + \frac{ 1}{y}\right)^{y }\left(1 + \frac{ 1}{y}\right)^{ 1}.
$$
Now you can recognize the limit as $y \to \infty$ as $e \times 1 = e$.
