If I have a real-valued function g defined on an interval [c,d] and I know that for $\epsilon > 0$, $\exists$ a $ \delta > 0$ s.t.
$\lvert g(x)-L\lvert$ < $\epsilon$, $ \forall x \in [c,d]$ where limit $L$ exists at point $p$ and $\lvert p - x \lvert < \delta$.
Why is it not necessarily the case that the difference between the supremum and infimum of $g(x)$ over the $x$-interval from $[p-\delta,p+\delta]$not necessarily < $2\epsilon$?
I see that f is not continuous or bounded. But closed [a,b] places a decent constrain on our max delta value, so I am having a hard time seeing how if the statement $\lvert g(x)-L\lvert$ < $\epsilon$ is true that the inequality would not hold.
Is the issue that g is not uniformly continuous, so if we change $\delta$ we may run into discontinuities that cause epsilon to fluctuate in size?
I'm not sure what I'm missing.