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Would I be right in thinking that: $x^ab^x\to0$ as $x\to \infty\,\,\forall a\in \mathbb R$ where $b\in [0,1)$? I think that $b^x$decays faster than the growth of $x^a$ but how might I prove that?

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Yes, but how to prove it? Do you know some result that might help? Have you already done some case (like $xe^{-x}$) so that you can examine the proof and see what may apply in this case? In summary, SHOW YOUR ATTEMPT, don't expect us to do it for you. – GEdgar Sep 18 '11 at 0:28
@GEdgar In fairness, the OP has explained what he thinks will happen and why he thinks that will be the case. It's not like it's a verbatim copy of a homework question. – Fly by Night Mar 6 '13 at 18:35

So its obvious for $a \le 0$ so take $a>0$. Then we have an indeterminate form and may use l'hopitals rule.

We change $x^a b^x$ into $\frac{x^a}{b^{-x}}$ and it is of the form $\frac{\infty}{\infty}$.

The strategy is to show it is of this form for some number of application of l'hopitals rule until it becomes $\frac{0}{\infty}\to 0$ and then the result will be proved.

Now you can prove by induction that $\lim_{x \to \infty} [\frac{d^n}{dx} b^{-x}] = \infty$ for all n.

We also know we can choose $n$ such that $a-n <0$ and if we differentiate $x^a$ n times we will have $\lim_{x \to \infty} \frac{d^n}{dx}x^a=0$.

So if we take a minimal n then all previous applications of l'hopitals rule were justified and the limit is indeed $0$.

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Take for example $\lim_{x \to \infty}x^ab^x = \lim_{x \to \infty} \frac{x^a}{B^x}$ where $B=\frac{1}{b}$, so if you log numerator and denominator you will see that the numerator is $O(\log x)$ and denominator $O(x)$, so $\lim_{x \to \infty} \frac{x^a}{B^x} = 0$

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I am afraid this is a circular approach to the question. – Siminore Aug 16 '12 at 11:06

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