# Limit for sequence with $e$-function

I need some help :)

$a,b \in \mathbb R$ and $a,b > 0$.

What is? $\lim_{x\to\infty} x^b e^{-ax}$

It should be $0$ but I'm not sure how to prove it... - EDIT: Without L-Hospital =(

Write it as $\lim_{x\to\infty} x^b e^{-ax}=\lim_{x\to\infty} e^{-ax+b\ln x}$. For large $x$ we have $|-ax|>|b\ln x|$, so only $-ax$ contributes. So we get $$\lim_{x\to\infty} x^b e^{-ax} =\lim_{x\to\infty} e^{-ax} =0$$
• Hello! I tried using this method with a similar seuqnece: $lim_{x\to 0} (x^a \cdot ln(x)^4) = lim_{x\to 0} (e^{aln(x)} * e^{ln(ln(x)^4)}$ Is this the correct beginning? You see, I once again have much problems.. – Vazrael Nov 29 '13 at 17:11
$$\lim_{x\rightarrow\infty}x^be^{-ax}=\lim_{x\rightarrow\infty}\frac{x^b}{e^{ax}}$$ Now use L'Hopital $\lfloor b+1 \rfloor$ times...
• what if $b$ is not integer? – the_candyman Nov 29 '13 at 16:52
• ok, but suppose that $b = 2.1$. Then you will have: $x^{2.1}, 2.1x^{1.1}, (2.1)(1.1)x^{0.1}, (2.1)(1.1)(0.1)x^{-0.9}, \cdots$ and so on... So you'll never reach $x^0$ – the_candyman Nov 29 '13 at 17:16
• right, but $x^{-.9}=\frac1{x^{.9}}$. and since you will have $lim_{x\rightarrow\infty}\frac1{Cx^{.9}e^{ax}}$, it will go to 0 right? – Eleven-Eleven Nov 29 '13 at 17:20