Limit iff Limits from Left and Right

Theorem
Let $f$ be a real function defined on an open interval $\left({a \, . \, . \, b}\right)$ except possibly at a point $c \in \left({a \, . \, . \, b}\right)$.

Then:
 * $f \left({x}\right) \to l$ as $x \to c$

iff:
 * $f \left({x}\right) \to l$ as $x \to c^-$, and
 * $f \left({x}\right) \to l$ as $x \to c^+$.

Proof

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

Then from the definition of the limit of a function, $\forall \epsilon > 0: \exists \delta > 0: 0 < \left\vert{x - c}\right\vert < \delta \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$.

So for any given $\epsilon$, there exists a $\delta$ such that $0 < \left\vert{x - c}\right\vert < \delta$ implies that $l - \epsilon < f \left({x}\right) < l + \epsilon$.

Now:

That is: $\forall \epsilon > 0: \exists \delta > 0$:


 * 1) $c - \delta < x < c \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$
 * 2) $c < x < c + \delta \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$

So given that particular value of $\epsilon$, we can find a value of $\delta$ such that the conditions for both:


 * 1) $f$ tending to the limit $l$ as $x$ tends to $c$ from the left, and
 * 2) $f$ tending to the limit $l$ as $x$ tends to $c$ from the right.

Thus $\displaystyle \lim_{x \to c} f \left({x}\right) = l \implies \lim_{x \to c^-} f \left({x}\right) = l$ and $\displaystyle \lim_{x \to c^+} f \left({x}\right) = l$.


 * Now let $f \left({x}\right) \to l$ as $x \to c^-$ and $f \left({x}\right) \to l$ as $x \to c^+$.

This means that:
 * 1) $\forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$, and
 * 2) $\forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$.

In the same manner as above, the conditions on $\delta$ give us that:

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

Thus:
 * $\displaystyle \lim_{x \to c^-} f \left({x}\right) = l$ and $\displaystyle \lim_{x \to c^+} f \left({x}\right) = l \implies \lim_{x \to c} f \left({x}\right) = l$

Hence the result.