Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
add comment

2 Answers 2

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 $0$.

share|improve this answer
add comment

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 $\lim_{x \to \infty} \frac{x^a}{B^x} = 0$

share|improve this answer
1  
I am afraid this is a circular approach to the question. –  Siminore Aug 16 '12 at 11:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.