Behaviour of Function Near Limit

Theorem
Let $$f$$ be a real function.

Let $$f \left({x}\right) \to l$$ as $$x \to \xi$$.

Then:
 * If $$l > 0$$, then $$\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: f \left({x}\right) > 0$$;
 * If $$l < 0$$, then $$\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: f \left({x}\right) < 0$$.

Proof
From the definition of limit of a function:


 * $$\forall \epsilon > 0: \exists \delta > 0: 0 < \left|{x - \xi}\right| < \delta \implies \left|{f \left({x}\right) - l}\right| < \epsilon$$


 * Let $$l > 0$$.

Since this is true for all $$\epsilon > 0$$, it is also true for $$\epsilon = l$$.

So let the value of $$\delta$$, for the above to be true, labelled $$h$$.

Then $$0 < \left|{x - \xi}\right| < h \implies \left|{f \left({x}\right) - l}\right| < l$$.

That is, $$\xi - h < x < \xi + h, x \ne \xi \implies 0 = l - l < f \left({x}\right) < l + l = 2l$$.

Hence $$\forall x: \xi - h < x < \xi + h, x \ne \xi: 0 < f \left({x}\right)$$.


 * Now let $$l < 0$$, and consider $$\epsilon = -l$$.

A similar thread of reasoning leads us to $$\xi - h < x < \xi + h, x \ne \xi \implies -2 l < f \left({x}\right) < 0$$, and hence the second result.

Note
If $$l = 0$$, neither conclusion may be drawn without further information - $$f \left({x}\right)$$ may stay either side of zero.