Prove $\lim\limits_{x\to 0^+} \frac{\ln x}{x} = -\infty$ 
Prove $\lim\limits_{x\to 0^+} \frac{\ln x}{x} = -\infty$

I've seen the following proof but I think it's invalid:
$$\lim\limits_{x\to 0^+} \frac{\ln x}{x} = \lim\limits_{x\to 0^+}\ln x \cdot \lim\limits_{x\to 0^+} \frac{1}{x} = -\infty \cdot \infty = -\infty$$
Is this proof valid? If not, I'd be glad for an alternative.  
Thanks.
 A: Let $y=\frac1x$, then as $x \rightarrow 0^+,y \rightarrow \infty$
$\lim\limits_{x\to 0^+} \frac{\ln x}{x}=\lim\limits_{y\to \infty} y \ln{\frac1y}=\lim\limits_{y\to \infty}- y \ln{y}$
As both $y$ and $\ln(y)$ increase to infinity, therfore $\lim\limits_{y\to \infty}- y \ln{y}=-\infty$
Hence $\lim\limits_{x\to 0^+} \frac{\ln x}{x}=-\infty$
A: set $\log n  = t, \ n = e^t$ and consider two cases:
$$
\lim_{t \to \infty} \frac{t}{e^t} = 0
$$
$$
\lim_{t \to - \infty} \frac{t}{e^t} = - \lim_{v \to \infty}e^v v  = - \lim_{v \to \infty} e ^{\log v + v} = - \infty
$$
to see that the one-sided limit doesn't exist (here set $-t = v$).
A: From an $M,\delta$ perspective, the proof is essentially correct.  Let's generalize slightly: 

Suppose $f$ and $g$ are two functions for which $\lim\limits_{x \to 0^+} f(x) = -\infty$, $\lim\limits_{x \to 0^+} g(x) = +\infty$.  Then, $\lim\limits_{x \to 0^+} f(x)\cdot g(x) = - \infty$.

Proof: Consider $f$ and $g$ with the desired properties. We will prove $\lim\limits_{x \to 0^+} f(x)\cdot g(x) = - \infty$ by showing that for any $M$, there exists a $\delta > 0$ such that for all $x$, $0 < x < \delta$, we have that $f(x)g(x) < M$. 
Consider real $M$, arbitrary. Since $\lim\limits_{x \to 0^+} f(x) = -\infty$, there exists a $\delta_1 > 0$ such that for all $x$ with $0 < x < \delta_1$, $f(x) < -1$. Also, since $\lim\limits_{x \to 0^+} g(x) = +\infty$, there exists a $\delta_M > 0$ such that for all $x$ with $0 < x < \delta_M$, $g(x) > \max\{-M, 0\}$.  Now, set $\delta = \min\{\delta_1, \delta_M\}$, and consider $x$ with $0 < x < \delta$.  There are two cases to consider.  If $M \ge 0$, then  $f(x)g(x) < 0 \le M$, as required.  Alternatively, if $M < 0$, then $-M > 0$, so we have  
$$f(x)\cdot g(x) < -1 \cdot g(x) < -1 \cdot -M \le M.$$
