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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.

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The reason is that $f$ may not be continuous at $x=p.$ That is, it may have any value there, since that doesn't affect the fact that it has a limiting value there. Clearly then, in such a case the difference in the max and min values can be as large as you please. For an example consider the function defined as $0$ for all $x\ne 0$ and which is $1$ at $0.$ Then the limiting value at $0$ is $0,$ and the difference in question on any interval $[-\delta,\delta],$ with $\delta>0$ is $1-0.$ Clearly there is some $\epsilon$ smaller than this when $\delta$ is sufficiently small.

The point is, that a function has a limiting value at a point and is defined there does not imply that its value there equals its limiting value. The situation you envisage (when the function is bounded over a sufficiently closed interval about $p$) is satisfied if in addition it is continuous at $p.$ That is, if it has a value that there that coincides with its limit there.

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    $\begingroup$ Clear as day. Thank you. $\endgroup$ – dingdong May 9 '19 at 8:56

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