2
$\begingroup$

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?

  • 0
    @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.2013-03-06

2 Answers 2

1

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 $0$.

0

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

  • 1
    I am afraid this is a circular approach to the question.2012-08-16