Order of convergence 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?
 A: 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 $0$.
A: 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 $\lim_{x \to \infty} \frac{x^a}{B^x} = 0$
