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 $$ \lim_{x\to\infty} x^b e^{-ax} =\lim_{x\to\infty} e^{-ax} =0 $$

  • $\begingroup$ 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.. $\endgroup$ – Vazrael Nov 29 '13 at 17:11
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    $\begingroup$ @K.L. sounds like another question... $\endgroup$ – draks ... Nov 29 '13 at 17:12

$$\lim_{x\rightarrow\infty}x^be^{-ax}=\lim_{x\rightarrow\infty}\frac{x^b}{e^{ax}}$$ Now use L'Hopital $\lfloor b+1 \rfloor$ times...

  • $\begingroup$ what if $b$ is not integer? $\endgroup$ – the_candyman Nov 29 '13 at 16:52
  • $\begingroup$ I know that the poster said no L'Hopital, but you can use L'Hopital whether or not b is an integer or not, right @the_candyman? Eventually, the numerator will run out of positive integers after between b and b+1 times you use L'Hopital and will end up in the denominator with a 1 in the numerator, so you'll still get the limit is 0. $\endgroup$ – Eleven-Eleven Nov 29 '13 at 17:09
  • $\begingroup$ 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 $x^0$ $\endgroup$ – the_candyman Nov 29 '13 at 17:16
  • $\begingroup$ 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? $\endgroup$ – Eleven-Eleven Nov 29 '13 at 17:20
  • $\begingroup$ yeah, that's right! :D $\endgroup$ – the_candyman Nov 29 '13 at 17:21

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