the limit exist iff one-sided limits exist and are equal

Question

Intuitively, this is definitely true since both one-sided limit meet at the same point, but I am struggling to proceed the proof. Sorry for not showing any attempt, but I have no idea of how to begin. Could you give any hint?

Thank you in advance.

Answer 1

If part. Given $\epsilon>0$, there are $\delta_{1},\delta_{2}>0$, if $c<x<c+\delta_{1}$, we have $|f(x)-L|<\epsilon$, and if $c-\delta_{2}<x<c$, then $|f(x)-L|<\epsilon$.

Now assume that $0<|x-c|<\delta=\min\{\delta_{1},\delta_{2}\}$, then either $c<x<c+\delta$ or $c-\delta<x<c$, in either case, one of $c<x<c+\delta_{1}$ or $c-\delta_{2}<x<c$ must be fulfilled, so we have $|f(x)-L|<\epsilon$.

Only if part. Given $\epsilon>0$, there is a $\delta>0$, if $0<|x-c|<\delta$, we have $|f(x)-L|<\epsilon$.

Now assume that $c<x<c+\delta$, then of course $0<|x-c|<\delta$, so we have $|f(x)-L|<\epsilon$, this shows that $\lim_{x\rightarrow c^{+}}f(x)=L$.

Can you do that for $\lim_{x\rightarrow c^{-}}f(x)=L$?