# 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.)

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There is either a typo in your title or a neologism... – Mariano Suárez-Alvarez Aug 7 '12 at 3:37

## 1 Answer

$\displaystyle \lim_{x\to \infty} \left(\frac {x}{e^x-1}\right)^2 e^x = \lim_{x\to \infty} \left(\frac {x^2}{e^x-1}\right)\left(\frac{ e^x}{e^x-1}\right)$ 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?

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Thanks! I forgot to make double derviatvie of the first one... – Asalph Tooks Aug 7 '12 at 3:48