Evaluate $\lim_{x\to 0^{+}}\frac{\ln(x)}{x}$ 
$$\lim_{x\to 0^{+}}\frac{\ln(x)}{x}$$

I know that $\lim_{x\to 0^{+}}\ln(x)=-\infty$ and $\lim_{x\to 0^{+}}\frac{1}{x}=\infty$ can I say anything about $-\infty\cdot\infty$ or it is intermediate of L'hopital?  
 A: We have, as $x \to  0^+$,
$$
\frac{\ln(x)}{x}=\frac{1}{x} \times \ln(x) \to  +\infty \times (-\infty)=-\infty.
$$
A: Another way to evaluate the limit is to note that $\log(x)\le x-1$ for all $x>0$, which I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality.  Therefore,
$$\begin{align}
\frac{\log(x)}{x}&\le \frac{x-1}{x}\\\\
&=1-\frac1x\\\\
& \to -\infty\,\,\left(\text{as}\,\,x\to 0^+\right)
\end{align}$$
A: Here is a formal proof which uses the definition of an infinite limit directly:
We have  $$\lim_{x\to 0^+}\frac{\ln x}{x}=-\infty$$ if for any $k>0$ one can find a $\delta>0$ such that for all $x\in(0,\delta)$ $$\frac{\ln x}{x}\leq-k.\quad (*)$$
Since $\ln x$ is concave it lies everywhere below the tangent at $x=1$, i.e. $$\ln x\leq x-1$$ for all $x>0$. Therefore $$\frac{\ln x}{x}\leq 1-\frac{1}{x}$$ for all $x>0$. 
Thus as long as$$1-\frac{1}{x}\leq -k,$$ then $(*)$ will also be true. So one can take $\delta=\frac{1}{1+k}$ (which satisfies the above inequality as an equality).
A: The expression is negative for $0<x<1$, so the limit is $-\infty$.
As you pointed out, $ln(x)$ tends to $-\infty$ and $\frac{1}{x}$ tends to $\infty$. So, the limit cannot be a finite real number.
You cannot apply L'Hospital here, because the limit of $\frac{1}{x}$ , $x$ tending to $0$ from above, does not exist either.
A: "I know that $\lim_\limits{x\to0+}\ln(x)=-\infty,$ and $\lim_\limits{x\to0+}1x= \infty$ can I say anything about $−\infty\cdot \infty$"
Lets parse this.  What dies $\lim_\limits{x\to0+}\ln(x)=-\infty$ really mean?
You learned the $\epsilon - \delta$ definition of limit, but it probably didn't stick the first time.  So, back to definitions.
$\forall M>0, \exists \delta > 0$ such that $0<x<\delta \implies \ln x < -M$
Pick an number that you think bounds $\ln x$ and I can find a value of $x$ in the neighborhood of $0$ that busts your bounds.
Now we add that we know something similar about $\frac 1x$ -- that I can find a value of $x$ such that $\frac 1x > M$
As long as $\frac 1x$ and $-\ln x$ are also greater than 1, then there exists an $x$ where $\frac {ln x}{x} < -M$
Mathing it up a little bit.
$\forall M>0, \exists \delta > 0$ such that $0<x<\delta \implies \frac {\ln x}{x} < -M$
$\lim_\limits{x\to 0+} \frac {\ln x}{x} = -\infty$
