TITLE: Prove directly from the definition that $\lim\limits_{x\to 0^+}\ln(x) = - \infty$
QUESTION [5 upvotes]: Prove directly from the definition that $\lim\limits_{x\to 0^+}\ln(x) = - \infty$

I tried to use the sequence definition of the limit, namely: 
Given $f: D \to R$  If $a$ is a limit point of $D\subseteq R$ and for all sequences $x_n \in D \to a$  $f(x_n) \to g,$ then 
$g = \lim\limits_{x \to a}f(x)$ 
So, first of all, let's take any sequence $x_n$ such that $x_n \to 0$. Now, consider the sequence $f(x_n) = \ln(x_n)$.  
As $x_n$ approaches zero, $\ln(x_n)$ will approach negative infinity (how can I prove this formally). What's more, we can infer that $x_n > \lim _{n\to \infty}\ln(x_n)$, and so 
$$\lim_{x \to 0^+} \ln(x) = - \infty$$
How can I improve my reasoning to make this proof clear and acceptable?

REPLY [4 votes]: Let $M<0$ be arbitrarily given. And let $n$ be a positive integer such that $n>-M/\log 2$ so that $n\log 2>-M$. If $\delta=1/2^{n}$ then for $0<x<\delta$ we have $$\log x<\log \delta=-n\log 2<M$$ and hence $\log x\to -\infty$ as $x\to 0^{+}$.

The above proof makes use of the following properties of logarithm function: $$\log (xy) =\log x+\log y, \forall x, y>0\text{ and } \log 2>0$$ These properties hold for logarithm to any base $b>1$ and therefore the limit result holds accordingly for logarithm to any base $b>1$.