# 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 $x \to c$
$f \left({x}\right) \to l$ as $x \to c^-$

and

$f \left({x}\right) \to l$ as $x \to c^+$

## Proof

### Necessary Condition

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:

 $\displaystyle$  $\displaystyle 0 < \left\vert{x - c}\right\vert < \delta$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle - \delta < -\left({x - c}\right) < 0$ $\displaystyle \lor$  $\displaystyle 0 < \left({x - c}\right) < \delta$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle c - \delta < x < c$ $\displaystyle \lor$  $\displaystyle c < x < c + \delta$

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

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

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

$(1): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the left

and :

$(2): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the right.

Thus:

$\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$

implies that:

$\displaystyle \lim_{x \mathop \to c^-} f \left({x}\right) = l$

and:

$\displaystyle \lim_{x \mathop \to c^+} f \left({x}\right) = l$

$\Box$

### Sufficient Condition

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:

$(1): \quad\forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$

and :

$(2): \quad\forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \left\vert{f \left({x}\right) - l}\right\vert < \epsilon$

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

 $\displaystyle$  $\displaystyle c - \delta < x < c$ $\displaystyle \land$  $\displaystyle c < x < c + \delta$ $\displaystyle \implies \ \$ $\displaystyle$  $\displaystyle 0 < \left\vert{x - c}\right\vert < \delta$

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:

$\displaystyle \lim_{x \mathop \to c^-} f \left({x}\right) = l$

and:

$\displaystyle \lim_{x \mathop \to c^+} f \left({x}\right) = l$

together imply that:

$\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$

Hence the result.

$\blacksquare$