Prove limit exists if and only if left and right limits exist and are equal Prove $\lim_{x\to a}f(x)=L \iff \lim_{x\to a^+}f(x)=L=\lim_{x\to a^-}f(x)$
I have no problem with the $(\Leftarrow)$ direction but I can't do it for the other direction. Proofs for both directions would be nice.  
 A: The definition for the limit of the function says:

If a limit $$\lim_{x \rightarrow a}f(x)=L$$ exists, then given any real number $\epsilon>0$, there exists another real number $\delta>0$ such that
  if $0< |x-a| <\delta $, then $|f(x)-L|<\epsilon$.

Now let us consider the first case, that is: $$\lim_{x \rightarrow a}f(x)=L \implies\lim_{x \rightarrow a^{+}}f(x)=L$$
For simplicity let us only consider a monotonously non-decreasing, continuous function. The proof will apply similarly to others.

We know that  $\forall$ $\epsilon >0$ $\exists$ $\delta>0$ $:$ $x<a+ \delta$ $\implies$ $f(x)<L+ \epsilon$ 

Now consider an $0<\epsilon_0<\epsilon$. Then there must exist a $\delta_0$  that satisfies 



*

*$\delta_0<\delta$, because remember that $f(x)$ is monotonously non decreasing and continuous in our interval

*$x<a+\delta_0$ $(=a^+ )$ $\implies f(x) < L+\epsilon_0<L+ \epsilon$ 
 

Hence, by the definition of the limit we have: $$\lim_{x \rightarrow a^+}f(x)=L$$
A similar proof can be used for the other side.
A: you can prove it by contradiction. assume that 1) the lim from the right = L and 2) the limit from the left = M, and assume that 3) the Lim = L. by definition 3) implies that the limit from the left is L which contradicts 2).
Repeat and assume that 3) the lim = M. this will contradict 1) this time.
therefore, by elimination the only possible value for the Lim = Lim from left = Lim from right.
