Prove that if $f$ is bounded and nondecreasing on $(a,b)$ then lim $f(x) $as $x$ approaches $b$ from the left exists.

I can see that $f(x_n)$ converges, since it is bounded and monotonic. Let me give some context and mention that we have dwelled mostly on the sequential definition of limits. Does $(x_n)$ need to be nondecreasing as well? I suppose I need to use sups and infs somehow, but could use a hint. This is not homework but a rather an optional practice problem to prepare for a test.

Solutions Collecting From Web of "Prove that if $f$ is bounded and nondecreasing on $(a,b)$ then lim $f(x) $as $x$ approaches $b$ from the left exists."

Hint: Choose any monotonically increasing sequence $x_n \to b$; then since $f$ is nondecreasing, $f(x_n)$ is a monotonically increasing and bounded sequence, with limit $L$.

Now take any sequence $y_n$ converging to $b$; can you convince yourself that $f(y_n) \to L$ as $y_n \to b$? Choosing a monotonically increasing subsequence $y_{n_k}$ may be of use to you.

Since $f$ is bounded we have $|f(x)|<M$ for some $M$ on $(a,b)$ Consider the set $\{f(x) : x<b\}$, this set is bounded above by $M$ thus it has a supremum $L$, that is $L=\sup\{f(x) : x<b\}$. I claim that $\displaystyle\lim_{x\to b^-}f(x)=L$. Let $\varepsilon>0$, by the definition of the $\sup$ there is a $y\in (a,b)$ such that $L-\varepsilon < f(y) \le L$ .Take $x \in (b-\delta,b)$ then $L-\varepsilon <f(y)\le f(x) \le L$ which means that $|f(x)-L|<\varepsilon$ when $0<b-x<\delta$, therefore $\displaystyle\lim_{x\to b^-}f(x)=L$.

Try finding the $\delta$.