Prove that: $\lim\limits_{x \to +\infty}\frac{(\log x)^b}{x^a} = 0$ and $\lim\limits_{x \to +\infty}\frac{x^b}{e^{ax}} = 0$ If $a > 0$  and $b > 0$, show that
$$\lim_{x \to +\infty}\frac{(\log x)^b}{x^a} = 0 \tag{1}$$
and
$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = 0 \tag{2}$$

Attempts:
$(1)$
Given that 
$$\log x = \frac{\ln x}{\ln 10}$$
Then
$$
\begin{align*}
\lim_{x \to +\infty}\frac{(\log x)^b}{x^a} 
&= \lim_{x \to +\infty}\left(\frac{\ln x}{\log 10}\right)^b\cdot \lim_{x \to +\infty}\frac{1}{x^a}\\
&= \lim_{x \to +\infty}\left(\frac{\ln x}{\log 10}\right)^b\cdot 0\\
&= 0
\end{align*}
$$
I'm not sure whether this is right. There is a Theorem a my textbook which says:

If $f(x)$ is an infinitesimal function as $x \to a$, and $g(x)$ is a bounded function, then $\lim_{x \to a}f(x)\cdot g(x)$ is an infinitesimal (i.e $= 0$).

The rightmost limit it is indeed and infinitesimal function, but it seems that the leftmost one is unbounded. Does the theorem not hold here?
$(2)$
$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = 0$$
Here I'm puzzled. So far I've done just this:
$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = \lim_{x \to +\infty}\frac{e^{b \ln x}}{e^{ax}} = ...$$

2: Assuming a and b are both positive and finite (second part wasn't mentioned, but I am assuming it). Expand the exponential term in the denominator as a sum.
$\frac{x^b}{e^{ax}}=\frac{x^b}{1+ax+\frac{1}{2}(ax)^2+\frac{1}{6}(ax)^3+....+\frac{1}{n!}(ax)^n}$ for $n\rightarrow\infty$. You can now use L'Hopital's rule on this in the limit where $x\rightarrow\infty$. In particular you will have some n that is large enough to cancel the $x^b$ term in the numerator, and still have some larger powers of x in the denominator. (e.g. you will be able to use L'Hopital's rule b times and still have a term dependent on x in the denominator.) Because $x\rightarrow\infty$, we are guaranteed that $(ax)>1$ for any finite $a$.
