Show that if $f$ is growing and bounded on $(a,b)$ than $\lim\limits_{x\searrow a}f$ and $\lim\limits_{x\nearrow b}f$ exist.
Since $f$ is bounded, there exists $\inf f$ and $\sup f$.
And since $f$ is growing, (1) $\forall x,y \in (a,b):x<y \Longrightarrow f(x) \le f(y)$
The definition of (2) $\sup f$ says $\forall x \in (a,b): f(x) \le \sup f$
Now we show:
$\forall \epsilon >0 \,\,\, \exists \delta>0 \,\,\,x \in (a,b):0<|x-b|< \delta \Longrightarrow |f(x)-\eta|<\epsilon$
We set $x:=b-\frac{\delta}{2}$ for any $\delta> 0$ and $0<|b-\frac{\delta}{2}-b|=|\frac{\delta}{2}]<\delta$ always holds.
Obviously $b-\frac{\delta}{2}\in (a,b)$ which means because of (2) $f(b-\frac{\delta}{2})\le \sup f$
If we additionally take $\sup f-\epsilon $ from the definition of the supremum it follows that:
$\sup f-\epsilon \in f((a,b)), \forall \epsilon >0$
Choosing $\delta$ in regards to any $\epsilon$ and since (1) gives us the inequation:
$\sup f-\epsilon < f(b-\frac{\delta}{2})\le \sup f$
Let $\sup f:= \eta$, we get $\eta-\epsilon < f(b-\frac{\delta}{2})\le \eta$
So $\forall \epsilon >0 \,\,\, \exists \delta>0 \,\,\,x \in (a,b):0<|x-b|< \delta \Longrightarrow |f(x)-\eta|<\epsilon$ indeed holds for the right side, which implies $\lim\limits_{x\nearrow b}f$ indeed exists!
For the second limit I would just do the same in the other direction :)
Could someone look over it and give some feedback?