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:
$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = \lim_{x \to +\infty}\frac{e^{b \ln x}}{e^{ax}} = ...$$

You must know something more, if you cannot use De l'Hospital. For instance
$$
\lim_{x \to +\infty} \frac{x^b}{e^x}=0 \quad\hbox{for every $b>0$}
$$
or
$$
\lim_{x \to +\infty} \frac{x}{e^{ax}}=0 \quad\hbox{for every $a>0$}.
$$
Indeed,
$$
\frac{x^b}{e^{ax}}=\left( \frac{x}{e^{\frac{a}{b}x}} \right)^b
$$
or
$$
\frac{x^b}{e^{ax}}=\left( \frac{x^{\frac{b}{a}}}{e^{x}} \right)^a
$$
For sure you cannot use the theorem in your book, since $\log x$ does not remain bounded as $x \to +\infty$.
