Real analysis : proving that $\lim_{x\to\ a} f(x)$ exists if $f$ is uniformly continuous on $(a,b)$.

Question

Defining $f$ to be a uniformly continuous function from $(a,b)$ to $R$, I want to prove that $\lim_{x\to\ a} f(x)$ and $\lim_{x\to\ b} f(x)$ exist.

I am saying that since $f$ is uniformly continuous, it can be extended to another uniform continuous function $g$ on $[a,b]$, so clearly $g(a)$ and $g(b)$ both exist. Can I just infer that $g(a) = \lim_{x\to\ a} f(x)$ and similarly for $g(b)$?

If not, what is the right way to prove this?

Answer 1

Choose any sequence of points $a < x_n \to b$. Let us show that $f(x_n)$ is a Cauchy sequence and therefore has a limit (due to completeness of $\mathbb{R}$). Given $\varepsilon > 0$, choose $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \varepsilon$ (possible due to uniform continuity of $f$) and $N$ such that $b-\delta < x_n < b$ for all $n > N$. Then $|f(x_k)-f(x_l)| < \varepsilon$ for all $k,l > N$.

It is easy to show that the limit $\lim f(x_n)$ does not depend on the choice of the sequence $(x_n)$. This is because any two sequences $(x_n),(y_n)$ convergent to $b$ can be merged into one: $x_1,y_1,x_2,y_2,\ldots$.

Answer 2

**hint **

use sequential characterization of the limit.

$x_n\to a^+ \implies (x_n) $ is Cauchy

$\implies (f(x_n))$ is Cauchy cause $f $ is u.continuous

$\implies lim_{n\to\infty}f (x_n) $ exists.