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

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

Now:

$$ $$ $$ $$ $$

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


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

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


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

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

$$ $$ $$

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

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

Hence the result.