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:

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

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.

$\blacksquare$

Note

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

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 8.15 \ (6)$