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