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 $\frac{\infty}{\infty}$? 
 A: No, you do not need to have any conclusions about $\ln(\infty/\infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$\lim_{x \to \infty} \ln \left(\frac{x - 1}{x}\right) = \lim_{x \to \infty} \ln \left(1 - \frac 1 x\right) = \ln 1 = 0$$
which is what you need.
A: HINT
Note that the inverse
$$\left(\frac{x-1}{x}\right)^x=\left(1-\frac{1}{x}\right)^x$$
A: 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 $e \times 1 = e$.
A: $$\lim\limits_{x\to \infty}\left(\frac{x}{x-1}\right)^x = \lim\limits_{y\to \infty}\left(\frac{y+1}{y}\right)^{y+1} = \lim\limits_{y\to \infty}\left(1 +\frac{1}{y}\right)\lim\limits_{y\to \infty}\left(1+\frac{1}{y}\right)^{y} = 1 \times e = e$$
A: Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$\lim_{x \to c} f(g(x)) = f \left( \lim_{x \to c} g(x) \right)$$
In your case,
$$\lim_{x \to \infty} \ln \left( \frac{x}{x-1} \right) = \ln \left( \lim_{x \to \infty} \left( \frac{x}{x-1} \right) \right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$\frac{x}{x-1} = \frac{x-1+1}{x-1} = 1 + \frac{1}{x-1}$$
