$\lim_{n \to \infty} \frac{\ln x^q}{x^p}$ not necessarily =0 for any $p>0$ and $q>0$ right? Is it true or false that$$\lim_{n \to \infty} \frac{\ln n^q}{n^p}=0$$ is not necessarily true for all $p>0$ and $q>0$?
I understand that $$\lim_{n \to \infty} \frac{(\ln n)^q}{n^p}=0$$ for all $p>0$ and $q>0$.
 A: Assuming you mean:
$$
\lim_{n \to \infty} \frac{\ln n^q}{n^p} = \lim_{n \to \infty} \frac{q \ln n}{n^p} = 0
$$
This for all $p > 0$
A: 
I understand that $$\lim_{n \to \infty} \frac{(\ln n)^q}{x^p}=0\tag{1}$$ for all $p>0$ and $q>0$.

If you know this to be true, then we can apply it here: for all $p, q > 0$  $$
\begin{align}\lim_{n \to \infty} \frac{\ln n^q}{x^p} 
& =  \lim_{n \to \infty} \frac {q\,\ln n}{n^p}\\ \\ 
& = q\lim_{n \to \infty} \frac {(\ln n)^1}{n^p} \tag{2} \\ \\
& = q\cdot 0 \tag{By (1), $q = 1 \implies (2) = q\cdot 0)$ } \\ \\
& = 0 \\ \\
\end{align}$$
A: Besides to other answers, note that the series : $$\sum_{n=1}^{\infty}\frac{(\ln n)^q}{n^p}$$ is convergent when $p>1, q\leq0$, since $$\lim_{n\to\infty}n^p\frac{(\ln n)^q}{n^p}<\infty$$
A: Edit: well finally it was $x=n$ and not $n=x$. The following is still true and it obviously implies the $n$ case. So I'll leave it like this.
The first ($\forall p,q>0$) follows from the second (with $q=1$), as $\ln x^q=q\ln x$. So let us prove the second. I'll do the general case for the sake of completeness, although we only need $q=1$.
Set $u:=x^p$, i.e. $x=u^{1/p}$. Then
$$
\frac{(\ln x)^q}{x^p}=\left(\frac{1}{p}\right)^q\frac{(\log u)^q}{u}.
$$
Applying L'Hospital $n$ times for some $n>q$ (e.g. $n=\lfloor q\rfloor+1$) yields
$$
\lim_{+\infty}\frac{(\log u)^q}{u}=q\lim_{+\infty}\frac{(\log u)^{q-1}}{u}=\ldots=q(q-1)\cdots(q-n+1)\lim_{+\infty}\frac{1}{u(\log u)^{n-q}}=0.
$$ 
Since $u\longrightarrow+\infty$ if and only if $x\longrightarrow +\infty$, you are done proving
$$
\lim_{+\infty}\frac{(\ln x)^q}{x^p}=0\qquad\forall p>0,q>0.
$$
