Knowing $a>1$ and $b>1$, how can we prove the limit of $\frac{x^a}{b^x}$ when $x$ goes to infinity? The limit is $0$, but I want to show it by the definition.
I know that, in order to do that, i must show that, given $\epsilon>0$, then there is a $M$ such that, for every $x>M$, we have $|\frac{x^a}{b^x}|>\epsilon$, but I was unable to develop the proof. Maybe there is a way by looking to the derivative and seeing how the function is decreasing.
 A: Taking logarithms, you could consider the function
$$f(x)=a \log (x)-x \log (b)$$ for which
$$f'(x)=\frac a x-\log(b) \qquad \text{and} \qquad f''(x)=-\frac a {x^2} \,\,< 0\,\,\forall x$$ The first derivative cancels at $x_*=\frac{a}{\log (b)}$ and the second derivative test shows that this is a maximum. Since, for $x>x_*$, $f'(x)<0$ then $f(x)$ is a decreasing function.
Since $\frac{x^a}{b^x}=e^{f(x)}$, then the same.
A: Write $x^a$ as $e^{a\log x}$ and $b^x$ as $e^{x\log b}.$ Then the function may be rewritten as $$e^{a\log x-x\log b}=e^{x\left(a\frac{\log x}{x}-\log b\right)},$$ which, as $x$ goes to $+\infty,$ goes to $e^{-\infty}=0,$ since $\log x/ x\to 0,$ and $\log b>0.$
A: $L =\log\frac{x^a}{b^x}= a \log x - x\log b \to - \infty$. Polynomial in $x$ grows faster than a log in $x$.
$\frac{x^a}{b^x}= e^L \to e^{-\infty}= 0 $
A: The limit is $0$ even if you take $a$ very large and $b$ very small e.g. $a=1000, b=1.1$ 
But you cannot prove this just by definition, it will be way too difficult, I believe. You need to base your proof on some previous theory/knowledge, which would give you which functions grow faster than which other functions.   
In particular if you know L'Hopital's rule, you should be fine. I mean from it you can derive that the limit is indeed zero.    
