Krdytkyu

I am going over a solution given to solving the follow limit,
$$lim_{xto infty}(frac{x}{x-1})^x$$
The solution continues as follows,

Consider raising the function to $e^{lncdots}$

We can find the limit as follows,
$$lim_{xto infty} x ln(frac{x}{x-1}) = lim_{xto infty} frac{ln(frac{x}{x-1})}{frac{1}{x}}$$

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 $frac{infty}{infty}$?

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

$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$

which is what you need.

HINT

Note that the inverse

$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$

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 $e times 1 = e$.

$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$

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

$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$