Proving $\left|\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}\right|$ tends to zero everyone, I'm trying to show that: $$\forall\, c\in\mathbb{R},\, d>0\,:\,\lim_{n\to\infty}\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}=0$$
Intuitively it's completely clear but formalizing it is proving to be a bit difficult.
I'd appreciate some help evaluating $\left|\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}\right|$ in a way that would lead to the result.
Thanks a lot.
 A: Substitute $x=e^t$ , then $$\lim_{x\to\infty}\frac{\left(\log\left(x\right)\right)^{c}}{x^{d}}=\lim_{t\to\infty}\frac{t^{c}}{e^{td}}$$
Now, Use L'Hopital's rule.
EDIT: On @Peter's suggestion,
Since functional limit $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\left(\log\left(x\right)\right)^{c}}{x^{d}}=0$$
Thus, for any sequence $\{a_n\}\to\infty $ $$\lim_{n\to\infty}f(a_n)=0$$
Choose $\{a_n\}=n$
A: If $c<0$ then $(\log(n))^c\to0$ and $n^d\to\infty$ as $n\to\infty$,whence the claim.
If $c=0$, also $n^d\to\infty$ gives us the desired result.
Assume $c>0$.
First show that 
$$\tag1 \lim_{x\to\infty}\frac{\ln x}{x}=0.$$
Then, substituting $x\leftarrow x^a$ with $a>0$ (so that $x\to+\infty\iff x^a\to+\infty$), this turns into
$$ \lim_{x\to+\infty}\frac{\ln x^a}{x^a}=\lim_{x\to\infty}\frac{a\ln x}{x^a}=0,$$
hence 
$$ \lim_{x\to+\infty}\frac{\ln x}{x^a}=0.$$
By taking $c$th powers, $c>0$, and letting $a=\frac dc$, we find
$$ \lim_{x\to+\infty}\frac{(\ln x)^c}{x^d}=0.$$
How to show $(1)$?
The inequality $e^y\ge 1+y$ for $y\in\mathbb R$ should be well-known. 
Then $e^y=(e^{y/2})^2\ge (1+\frac y2)^2=1+y+\frac 14y^2$ for $y\ge -2$ so that substituting $x\leftarrow e^y$, the claim in $(1)$  follows from
$$0\le \lim_{y\to+\infty}\frac y{e^y}\le \lim_{y\to+\infty}\frac y{1+y+\frac14y^2}\le \lim_{y\to+\infty}\frac y{\frac14y^2}=0.$$
A: One way is to use L'hopital's rule on:
\begin{equation}
f(x)=\frac{(\log x)^c}{x^d}
\end{equation}
and conclude the result for $f(n)$.
