Q:

limit of exponential function and applying l'Hospital's rule

The eqaution goes the following:
$$\lim_{x\to \infty} \left(\frac {x}{e^x-1}\right)^2 e^x = \ ? $$
The first question is I tried to use l'Hopistal's rule, but unsure whether this is the right approach, as the limit goes to the infinity. (and this way did not produced a right answer, which seems to be zero.)
And what is the right way for solving this? (I know what this is when the limit $x$ goes to $0$, which is $1$, but unsure of this case.)

A:

$\displaystyle \lim_{x\to \infty} \left(\frac {x}{e^x-1}\right)^2 e^x = \lim_{x\to \infty} \left(\frac {x^2}{e^x-1}\right)\left(\frac{ e^x}{e^x-1}\right) $ and  both the terms in the product  are now easy to evaluate, $\displaystyle\lim_{x\to \infty} \left(\frac {x^2}{e^x-1}\right)=0$ and $\displaystyle\lim_{x\to \infty} \left(\frac{ e^x}{e^x-1}\right)=1$, by applying L'hopital's rule on both these limits.
Can you now see why the limit is zero?

