Show that for any k, $\displaystyle \lim_{x\to\infty}\frac{x^{k}}{e^{x}}\to0$.

I'm thinking that this has something to do with l'Hopital's Rule, showing that $\displaystyle x^{k} > e^{x}$, not sure where to start though?

Thank in advance

Printable View

- Mar 1st 2010, 07:39 AMcraigLimits Problem
Show that for any k, $\displaystyle \lim_{x\to\infty}\frac{x^{k}}{e^{x}}\to0$.

I'm thinking that this has something to do with l'Hopital's Rule, showing that $\displaystyle x^{k} > e^{x}$, not sure where to start though?

Thank in advance - Mar 1st 2010, 08:24 AMchisigma
Applying k times L'Hopital's Rule You obtain...

$\displaystyle \lim_{x \rightarrow \infty} \frac{x^{k}}{e^{x}} = \lim_{x \rightarrow \infty} \frac{k!}{e^{x}} = 0$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$