If $f : [a,b]\to\Bbb R$ is monotonic on $[a,b]$, then $\lim_{x\to b^-}f(x)$ exists and it is equal to $\sup\{f(x)\,:\,x\in[a,b)\}$ 
If $f : [a,b] \rightarrow \mathbb{R}$ is monotonic on $[a,b]$, then the left limit $\lim_{x\to b^-} f(x)$ exists and it is equal to $\sup\{f(x)\,:\,x\in[a,b)\}$

I know that the limit exists but I need someone to confirm whether it is equal to $\sup\{f(x)\,:\,x\in[a,b)\}$
Thank you
 A: Let $\epsilon>0$ given and $M=\sup E$ with $E=\{f(x),x\in [a,b)\}$.
then
$\exists y_0\in E \;:\; M-\epsilon <y_0\leq M$
$\exists x_0\in [a,b)\;:\: M-\epsilon<f(x_0)\leq M$
$$\implies \forall x\in (x_0,b) \;\; $$
$$M-\epsilon<f(x_0)\leq f(x)\leq M$$
put $x_0=b-\eta$
then
$$\forall x\in(b-\eta,b)\;\; |f(x)-M|<\epsilon$$
$$\lim_{x\to b^-}=M$$.
A: Let $L = \sup\{f(x) : x∈[a,b)\}$. 
For arbitrary $\epsilon \in (0, \frac{|b-a|}{2})$ consider $(L - \epsilon, L)$. By assumption, the preimage of $(L - \epsilon, L)$ must be nonempty, otherwise, since $f$ clearly cannot attain its supremum $L$, $L-\epsilon$ would be the $\sup$ instead. Thus, for some $d \in [a,b)$, $f(d) \in (L - \epsilon, L)$. 
We need to show for some $\delta>0$ $f[(b - \delta, b)] \subset (L - \epsilon, L)$. This is equivalent to the assertion that 
$$\lim_{x \to b^{-}} f(x) = L$$
We can prove this by contradiction. 
Suppose to the contrary that for all $\delta >0$ there is some $c_{\delta} \in (b - \delta, b)$ with $f(c_{\delta}) \notin (L - \epsilon, L)$. Then we must have $f(c_{\delta}) \leq L - \epsilon$. Note that such a $c_{\delta}$ exists regardless of our choice of $\delta$.  
Thus, if we choose $\delta_0$ such that $a< d < b - \delta_0 < c_{\delta_0} < b$ we have $f(d) \leq f(c_{\delta_0}) \leq L - \epsilon$ since $f$ is increasing. This contradicts the fact that $f(d) \in (L - \epsilon, L)$.
