Show $\lim \limits_{x \rightarrow c_{-}} f(x) \neq \lim \limits_{x \rightarrow c_{+}} f(x)$ imply $f$ is discontinuous at $c$ 
How to show $\lim \limits_{x \rightarrow c_{-}} f(x) \neq \lim \limits_{x \rightarrow c_{+}} f(x)$ imply $f: \mathbb R \rightarrow \mathbb R$ is discontinuous at $c$ ?

I know that $f$ cannot have two different limits at $c$, since this is causing a contradiction.
Also if I draw $f$ under the given assumption I see there is a jump on the graph, which intuitively imply discontinuity.
How can I prove this is numbers ?
 A: *

*Write the definition of what $$\lim_{x\to c_-} f(x) = a$$ means. 

*Do the same with the other limit. 

*write the definition of the continuity of $f$ at $c$. 

*Compare what you have and comment on it. 


I cannot promise you that you will get the answer right away, but until you do these things, it is very hard for us to help you.
A: Let:
$$\lim \limits_{x \rightarrow c_{-}} f(x)= M$$
$$ \lim \limits_{x \rightarrow c_{+}} f(x) = L$$
Without loss of generality let $M > L$ (otherwise repeat the same arguments, etc.)
These statements mean that there exists a $\delta_1, \delta_2$ so that for any $\epsilon$ the following holds:
$$c < x < c + \delta_1 \implies |f(x) - M| \lt \epsilon$$
$$c - \delta_2 \ < x < c  \implies |f(x) - L| \lt \epsilon$$
Take $\delta = \min(\delta_1, \delta_2)$ and $\epsilon = \frac{M-L}{2}$.
Combining the above statements, for $0 < |x-c| < \delta$, it must be that both $|f(x) - M| < \frac{M-L}{2}$ and that $|f(x) - L| < \frac{M-L}{2}$.
Can you arrive at a contradiction from this? It is essentially the same from here as the proof that a function cannot have two different limits at a point.
A: Perhaps this perspective will help you.
The statement you seek to prove is equivalent to (its contrapositive):


*

*If $f$ is continuous at $c$, then $ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x).$


But that comes from the definition of continuity of $f$ at $c$.
