How to prove that if the one sided limits are equal the general limit is that value?

My teacher proposed this question as a challenge proof to do on our own and I can't seem to get it. Was Wondering if anyone could give me a hint or help me on the process to completing it.

Let $a \in \mathbb{R}$. Let $f$ be a function defined, at least, on an interval centred at $a$, except possibly at $a$. Let $L \in \mathbb{R}$.

If $\lim\limits_{ x→a-} f(x) = \lim\limits_{ x→a+} f(x) = L$ then $\lim\limits_{ x→a} f(x) $ = $L$.

We are supposed to prove this using the "delta-epsilon" proof style. Thanks in advance!!

According to assumptions you have:

$$\eqalign{ & \mathop {\lim }\limits_{x \to {a^ + }} f(x) = L \cr & \exists L:\left( {\forall \varepsilon > 0,\exists {\delta _1} > 0:\left( {\forall x,0 < x - a < {\delta _1} \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right) \cr & \cr & \mathop {\lim }\limits_{x \to {a^ - }} f(x) = L \cr & \exists L:\left( {\forall \varepsilon > 0,\exists {\delta _2} > 0:\left( {\forall x,0 < a - x < {\delta _2} \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right) \cr} $$

Now just combine the two above relations by considering $\delta = \min \left\{ {{\delta _1},{\delta _2}} \right\}$ to get

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

which means

$$\mathop {\lim }\limits_{x \to a} f(x) = L$$

Start by writing your hypothesis in terms of delta-epsilon, and what you want to prove, that is

$$ \lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=L $$

Implies that given $\epsilon >0$ there exist $\delta_1,\delta_2 >0$ such that if:

$x \in (a-\delta_1,a)$ then $|f(x)-f(a)| < \epsilon$ (which is the formal definition of $\lim_{x\to a^{-}}f(x)=L$)

and if

$x \in (a,a+\delta_2)$ then $|f(x)-f(a)| < \epsilon$ (which is the formal definition of $\lim_{x\to a^{+}}f(x)=L$)

You want to find a $\delta >0$ such that for any $x \in (a-\delta,a+\delta)$ you have that $|f(x)-f(a)|<\epsilon$

What $\delta>0$ are you going to choose to ensure that the last statement is true?