Check if $\ln(x), x > 0$ is uniformly continuous
My only idea on solving this was to use the definition of uniform continuity. Namely, I need to show that for all $\epsilon >0$ there exists a $\delta >0$ such that for all $x_1, x_2$ in the domain of the function $|x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$
I have never before proved anything like this and I have no idea whether or not this function is uniformly continuous. Anyway, looking at the tremendous slope close to zero I am inclined to believe that this function is not uniformly continuous.
I want to show that $$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$
And so let's pick $\epsilon = 1$
I want to show that we can pick such $x_1, x_2$ that $|\ln(x_1)- \ln(x_2)| > 1 \Rightarrow |x_1 - x_2| > \delta$ for all $\delta$.
We know that $x_1 \ge ex_2$.
And now, let's take an arbitrary $\delta$, such that
$$|x_1 - x_2| \ge \delta$$
Now, we can simply pick
$$x_1 = ex_2 + \delta$$
$$x_2 = 1$$
And it works. And so the function is not uniformly continuous.
I know that my solution is very chaotic and confusing,is there an easier way to tackle this?