Is uniform continuity needed here…?

Question

I found this problem, but I don't think uniform continuity is required.

If $f$ is uniformly continuous on $(0,1)$, show $\lim_{x \to 0^{+}} f(x)$ exists.

Doesn't this just fall from $f$ being continuous?

Answer 1

No, plain continuity is not sufficient. Consider $f(x)=1/x$. This is defined and continuous on the whole interval. If infinite limits are considered to exist, $f(x)=\sin(1/x)$ is another counterexample.

Answer 2

First, functions like $1/x$ or $\sin(1/x)$ show that continuity is not enough.

On the other hand, uniform continuity does suffice. Indeed, let $a_n=f(1/n)$; we'll show that $a_n$ is a Cauchy sequence, and its limit will be the limit of $f(x)$ when $x\to0^+$.

Given any $\epsilon>0$ and $x\in(0,1)$ there is $\delta>0$ such that $|\xi-x|<\delta\to|f(\xi)-f(x)|<\epsilon/2$. Since $1/n$ is a Cauchy sequence, there exists some $n_0$ such that $q>p\ge n_0\to1/p-1/q<\delta$. That is, $$q>p\ge n_0\to|f(1/p)-f(1/q)|<\epsilon/2$$

We have proven that $a_n$ is a Cauchy sequence and hence converges. Let $a=\lim a_n$. For that $\delta$ formerly found, pick $x\in(0,\delta)$, and $n>1/\delta$ such that $|f(1/n)-a|<\epsilon/2$.

$$|f(x)-a|\le|f(x)-f(1/n)|+|f(1/n)-a|<\epsilon/2+\epsilon/2=\epsilon$$