Prove directly from the definition that $\lim\limits_{x\to 0^+}\ln(x) = - \infty$ 
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?
 A: Note that
$$\lim_{x \to 0+} \ln x = -\lim_{x \to 0+} \ln \frac{1}{x},$$
and for $0 < x < 1$
$$\ln \frac{1}{x} = \int_1^{1/x}\frac{dt}{t} > \sum_{k=1}^{\lfloor 1/x \rfloor-1}\int_k^{k+1}\frac{dt}{t} > \sum_{k=1}^{\lfloor 1/x \rfloor-1}\frac{1}{k+1}.$$
Now use the fact that the harmonic series diverges to $+\infty$.
A: 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 $b>1$.
A: Usually $\ln x$ is defined by $\ln x = \int_1^xt^{-1}\,dt$. If $x_n \to 0^+$ as $n\to+\infty$, then consider the sequence $\int_1^{x_n}t^{-1}\,dt$. Eventually $x_n < 1$, so for all $n$ sufficiently large we can write $\int_1^{x_n}t^{-1}\,dt = -\int_{x_n}^1t^{-1}\,dt$ with $\int_{x_n}^1t^{-1}\,dt > 0$.
The integral $\int_0^1 t^{-1}\,dt$ diverges in the sense that for any sequence $x_n\to 0^+$, we have $\int_{x_n}^1t^{-1}\,dt \to +\infty$ as $n\to+\infty$. To check this, compare the integrals of $t^{-1}$ and $f_n(t) = \min(n,t^{-1})$; note that $\int_0^1 f_n(t)\,dt$ is finite for every $n$, $\int_0^1 f_n(t)\,dt \le \int_0^1 t^{-1}\,dt$ since $f_n(t) \le t^{-1}$ for every $t$, and $\int_0^1f_n(t)\,dt \to +\infty$ as $n\to+\infty$.
Thus for all sufficiently large $n$, $\ln x_n = -\int_{x_n}^1t^{-1}\,dt \to -\infty$ as $n\to+\infty$ since $\int_{x_n}^1 t^{-1}\,dt \to +\infty$ as $n\to+\infty$.
A: you can prove directly by the definition, if you fix $$M<0$$ you can always find a value $\epsilon >0$ $$x=0+\epsilon$$ such that $$\log{x} < M$$ infact it is sufficient to select $\epsilon$ such that: $$\log{\epsilon} < M$$ that is$$\epsilon < e^M \ \square$$
A: This follows from
$\ln(x)+\ln(1/x) = 1$
and
$\lim_{x \to \infty} \ln(x)
= \infty$.
The latter follows from
$\ln(x)
=\int_1^x \frac{dt}{t}$
and the divergence of
$\sum \frac1{n}$
in the elementary form
$\sum_{n=1}^{2^m} \frac1{n}
\ge m/2
$.
