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 $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: \map f x > 0$
 * If $l < 0$, then $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: \map f x < 0$

Proof
From the definition of limit of a function:


 * $\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - \xi} < \delta \implies \size {\map f x - l} < \epsilon$

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:
 * $0 < \size {x - \xi} < h \implies \size {\map f x - l} < l$

That is:
 * $\xi - h < x < \xi + h, x \ne \xi \implies 0 = l - l < \map f x < l + l = 2 l$

Hence:
 * $\forall x: \xi - h < x < \xi + h, x \ne \xi: 0 < \map f x$

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

A similar thread of reasoning leads us to:
 * $\xi - h < x < \xi + h, x \ne \xi \implies -2 l < \map f x < 0$

and hence the second result.

Note
If $l = 0$, neither conclusion may be drawn without further information - $\map f x$ may stay either side of zero.