Prove that $\ln x$ is not uniformly continuous on $(0,1]$

Question

Prove that $\ln x$ is not uniformly continuous on $(0,1]$.

Got all confused on how to prove this kind of things... Any hints would be great!

Answer 1

Is obvious right away that $\ln x$ is not uniformly continous on $\left(0,1\right]$ because it has no limit as $x\to 0$, and it is known that any uniformly continous function on a bounded interval has limits as $x$ gets close to one of the extremities.

To prove it directly, use the definition of a uniformly continous function:

$f$ is uniformly continous when, given any $\varepsilon >0$, there exists $\delta>0$ such that, for all $x$ and $y$ in $\left(0,1\right]$: $$\text{if $x$ and $y$ are such that } \left|x-y\right|<\delta \text{, then } \left|f\left(x\right)-f\left(y\right)\right|<\varepsilon \text{ .}$$

Well, let $\varepsilon$ be a positive real number, and suppose that there existed $\delta$ satisfying the above condition. Take any $x\in\left(0,\delta\right]$. Then, for any $y$ in $\left(0,x+\delta\right)$ we would have $\left|\ln x - \ln y\right|<\varepsilon$, which means that $\ln y \in \left(\ln x - \varepsilon, \ln x + \varepsilon\right)$, and in particular, $\forall y\in \left(0,x+\delta\right), \ln y > \ln x - \varepsilon$. Well, that's clearly an absurd, because $\ln y$ tends to $-\infty$ as $y\to 0$. Thus $\ln$ cannot be uniformly continous in $\left(0,1\right]$

Answer 2

A function is not uniformly continuous if there exist two sequences $x_k, y_k$ such that $$ x_k - y_k \to 0 \qquad\text{but not}\qquad f(x_k)-f(y_k) \to 0. $$ This can be obtained directly by negating the definition of uniform continuity.

So if $\lim_{x\to x_0} f(x) = -\infty$ then such sequences can be found (take any sequence $x_k\to x_0$ and a sequence $y_k \to x_0$ such that $|f(y_k)-f(x_k)|\ge 1$) and the function is not uniformly continuous.