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!!
 A: 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
$$\mathop {\lim }\limits_{x \to a} f(x) = L$$
A: 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?
