Limit $\lim_{x \to \infty} \frac{x}{\log(x)}$. I want to find 

$$\lim_{x \to \infty} \frac{x}{\log(x)}$$

I can see that it will be infinity as $x$ increases too fast then logx.
But how can I find it mathematically?
I tried using lhospital rule but as definition of lhospital rule says that it is allowed to use if answer of limit exists, but here answer is infinity so I think that I can not use it.
Please correct me in case I am wrong.
Thank you.
Edit1:
Can we use lhospita here or not. Will using lhospital will be valid or not.
 A: $$x=\exp(\log x)=1+\log x +\frac{(\log x)^2}2+\cdots>\frac{(\log x)^2}2$$
if $x>1$ and so
$$\frac x{\log x}>\frac{\log x}2$$
etc.
A: By substitution
$$\frac x{\log x}=\frac {e^{\log x}}{\log x}=$$
set $\log x = y \to +\infty$
$$=\frac {e^{y}}{y}\stackrel{\text{definitively}}>\frac {y^2}{y}=y\to+\infty$$
You can also use l'Hospital:

$$\frac x{\log x} \stackrel{\text{Hospital}}\implies\frac 1{1/x}=x\to+\infty$$

A: Suppose $x\gt 4$ then $$\frac {x^2}{\log {x^2}}=\frac x2\cdot\frac x{\log x}\gt2\cdot \frac x{\log x}$$You can use this to show that the function is unbounded as $x\to \infty$
A: $x \rightarrow +\infty.$
Let $x:=e^y; $
exponential fct. is bijective with $D= \mathbb{R}$, $R= (0,\infty).$
Consider for $y \gt 0$:
$F(y): =\dfrac{e^y}{y} = \dfrac{1+y + y^2/2! +...}{y}$
$\gt y/2.$
$\lim_{y \rightarrow \infty} F(y) =\infty.$
A: May I suggest this is a simple high school limit?
One learns in high school that
$$\lim_{x\to \infty}\frac{\log x}x=0^+$$
so the answer is obvious by the general rules for limits.
This standard limit is obtained  proving that
$$\forall x>1,\;0<\log x<2\sqrt x\quad\text{whence}\quad 0<\frac{\log x}x <\frac1{\sqrt x}.$$
