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:

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

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:

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

and so:

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

• 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:

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

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:

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

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.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 8.4$