Prove that $\ln(x)$ is not uniformly continuous on $(0, \infty)$ 
Prove that $\ln(x)$ is not uniformly continuous on $(0, \infty)$

I have written a proof of this but I am not sure whether or not it's formally correct.
Assume that this function is uniformly continuous. Let's put $\epsilon = 1$. Now, choose $x$ and $y$ so, that $|x-y| < \delta$. Now, set $x= \delta$ and $y = \frac{\delta}{1000}$. Now, by definition of uniform continuity:
$$|\ln(x) - \ln(y)| < 1 \iff |\ln(\frac{x}{y})| < 1 \iff \ln(1000) < 1$$
Contradiction. 
Could you tell me whether the method I used is correct?
 A: You're not quite there yet, be a bit more precise:
Suppose $f(x) = \ln(x)$ is uniformly continuous. Then, taking $\varepsilon=1$ there must be some $\delta > 0$ such that 
$$\forall x,y \in (0,\infty): |x-y| < \delta \rightarrow |f(x) - f(y)| = |\ln(\frac{x}{y})| < 1$$
Then define $x = \delta>0$ and $y = \frac{\delta}{100}>0$ and note that this obeys $|x-y| = \delta\frac{99}{100} < \delta$ while $\frac{x}{y} = \frac{\delta}{\frac{\delta}{100}} = 100$ and $\ln(100)> 1$ contradiction, so no such $\delta$ can exist.
A: With $x_n=\frac {2}{n} $ and $y_n=\frac {1}{n} $, we have
$$\lim_{n\to+\infty}(x_n-y_n)=0$$ but
$$\lim_{n\to+\infty}(\ln (x_n)-\ln (y_n))=\ln (2)\ne 0$$ thus
$x\mapsto \ln (x) $ is not uniformly continuous at $(0,+\infty) $.
A: Your proof is correct but could be improved. What about the following changes? Assume that this function is uniformly continuous. Let's put $ϵ=1$. For a given $ \delta >0$, set $x=δ$ and $y=δ/1000$. Now
$|\ln(x)−\ln(y)|=|\ln(x/y)|=\ln(1000)>1$.
A: Another attempt;
Let $\epsilon=1$ be given.
Assume there is a $\delta >0$ such that 
$ |x-y| \lt \delta$, with $ x,y \in (0,\infty)$, implies
$|f(x)-f(y) |\lt 1$.
Choose $x_n = e^{-n}$, $n\in \mathbb{N}$.
$ \lim_{n \rightarrow \infty}x_n=0$.
Hence $x_n$ is a Cauchy sequence:
For $\delta $ there is a $m_0$, such that for $n,m \gt n_0$
$|x_n-x_m| \lt \delta.$
Choose $m=m_0$, and consider:
$|\log x_n - \log x_{m_0}| = $
$|-n +m_0| \lt 1.$
Choose n large enough to get a contradiction.
