Behaviour of Function Near Limit
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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.
$\blacksquare$
Note
If $l = 0$, neither conclusion may be drawn without further information - $\map f x$ may stay either side of zero.
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 8.15 \ (6)$