# showing that $\lim_{x\to b^-} f(x)$ exists.

please I require help in showing that if $f$ is uniformly continuous on a bounded interval $(a,b)$, then $\lim_{x\to b^-} f(x)$ exists.

edit:

$f$ is uniformly continuous on $(a,b)$ implies that for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $|f(x) -f(y)| \lt \epsilon$ whenever $|x-y|\lt \delta$ for every $x,y$.

let $x_n$ and $y_n$ be sequences in $(a,b)$ that converge to $b$. Then there is a natural number $N$ such that for every $n \geq N$, $0<b-x_n< \delta/2$ and $0<b-y_n< \delta/2$ so that $$|x_n-y_n| \leq |x_n -b| + |y_n -b| \lt \delta.$$ So, $|f(x_n) -f(y_n)| \lt \epsilon$ and this implies that $\lim f(x_n) = \lim f(y_n)$.

#### Solutions Collecting From Web of "showing that $\lim_{x\to b^-} f(x)$ exists."

Expanding azarel’s hint:

1) Show that a uniformly continuous function maps Cauchy sequences to Cauchy sequences. This isn’t hard to do (and is a very useful property), it follows almost directly from the relevant definitions.

2) Take a fixed sequence $(x_n)$ in $(a,b)$ converging to $b$. By the first step, $\bigl(f(x_n)\bigr)$ is Cauchy and thus converges to some number $\alpha$.

3) Now suppose that $\lim\limits_{x\rightarrow b^-} f(x)\ne\alpha$. Then you can find an $\epsilon>0$ and a sequence $(y_n)$ in $(a,b)$ converging to $b$ such that for all $n$, $|f(y_n)-\alpha|>\epsilon$.

4) Consider what we have: $(x_n)$ and $(y_n)$ are both sequences in $(a,b)$ converging to $b$. On one hand we have that $|x_n-y_n|$ can be made as small as you like. But on the other hand, for large $n$, the quantity $|f(x_n)-f(y_n)|\approx|\alpha-f(y_n)|$ is big. Could this occur for a function that is uniformly continuous on $(a,b)$?

$\bf Hint:$ Show that if $(x_n)\subseteq (a,b)$ converges to $b$ then $(f(x_n))$ is a Cauchy sequence and then show that the limit does not depend on the chosen sequence.