Limit iff Limits from Left and Right

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Theorem

Let $f$ be a real function defined on an open interval $\openint a b$ except possibly at a point $c \in \openint a b$.


Then:

$\map f x \to l$ as $x \to c$

if and only if:

$\map f x \to l$ as $x \to c^-$

and

$\map f x \to l$ as $x \to c^+$


Proof

Necessary Condition

Let $\map f x \to l$ as $x \to c$.

Then from the definition of the limit of a function:

$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$

So for any given $\epsilon$, there exists a $\delta$ such that:

$0 < \size {x - c} < \delta$

implies that:

$l - \epsilon < \map f x < l + \epsilon$

Now:

\(\displaystyle \) \(\) \(\displaystyle 0 < \size {x - c} < \delta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle - \delta < -\paren {x - c} < 0\)
\(\displaystyle \lor\) \(\) \(\displaystyle 0 < \paren {x - c} < \delta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle c - \delta < x < c\)
\(\displaystyle \lor\) \(\) \(\displaystyle c < x < c + \delta\)


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

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


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:

$\displaystyle \lim_{x \mathop \to c} \map f x = l$

implies that:

$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$

and:

$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$

$\Box$


Sufficient Condition

Let $\map f x \to l$ as $x \to c^-$ and $\map f x \to l$ as $x \to c^+$.

This means that:

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

and :

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

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

\(\displaystyle \) \(\) \(\displaystyle c - \delta < x < c\)
\(\displaystyle \land\) \(\) \(\displaystyle c < x < c + \delta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle 0 < \size {x - c} < \delta\)


So:

$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$


Thus:

$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$

and:

$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$

together imply that:

$\displaystyle \lim_{x \mathop \to c} \map f x = l$

Hence the result.

$\blacksquare$


Sources