$f$ is a monotonic increasing function. Prove $\lim_{x\to x_{{0}^{+}}} f(x)= \inf_{x>x_0} f(x)$

Question

$f$ is a monotonic increasing function. Prove that

$\lim_{x\to x_{{0}^{+}}} f(x)= \inf_{x>x_0} f(x)$

and that

$\lim_{x\to x_{{0}^{+}}} f(x) \le f(x_0) \le \lim_{x\to x_{{0}^{+}}} f(x)$

This is clearly true but I'm struggling with a formal proof. Any help would be appreciated.

Answer 1

Hint for the first part: We know that the infimum must exist. So let $$A = \inf_{x>x_{0}} f(x)$$ For any $\epsilon>0$, we can always find a $y$ such that $A + \epsilon > f(y)$, so that $|f(y)-A|<\epsilon$.

We can now say that if $x_{0}<x<y$, so if $|x-x_{0}|<|y-x_{0}|=\delta$, then $|f(x)-A|<\epsilon$.

Answer 2

As $f$ is increasing for all $x>x_0\quad f(x)>\alpha=f(x_0)$ (the lower bound).

Using the definition of the limit, you have to show $\forall \varepsilon>0, \exists c\in \mathcal{D}_f\quad$ s.t. $f(c)<\alpha+\varepsilon$