Answer
$\lim\limits_{x \to a}f(x) = L~~$ if and only if $~~\lim\limits_{x \to a^-}f(x) = L = \lim\limits_{x \to a^+}f(x)$
Work Step by Step
Assume that $\lim\limits_{x \to a}f(x) = L$
Then, according to Definition 2, for any $\epsilon \gt 0$, there is a $\delta \gt 0$ such that if $0 \lt \vert x-a \vert \lt \delta~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
Then:
If $a-\delta \lt x \lt a~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
If $a \lt x \lt a+\delta~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
Then, according to Definition 3 and Definition 4:
$\lim\limits_{x \to a^-}f(x) = L = \lim\limits_{x \to a^+}f(x)$
Now assume that $\lim\limits_{x \to a^-}f(x) = L = \lim\limits_{x \to a^+}f(x)$
Then, according to Definition 3 and Definition 4, for any $\epsilon \gt 0$, there is a $\delta_1 \gt 0$ and $\delta_2 \gt 0$ such that:
If $a-\delta_1 \lt x \lt a~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
If $a \lt x \lt a+\delta_2~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
Let $\delta = min\{\delta_1, \delta_2\}$
Then:
If $~~0 \lt \vert x-a \vert \lt \delta~~$ then $~~\vert f(x)-L\vert \lt \epsilon$
According to Definition 2:
$\lim\limits_{x \to a}f(x) = L$
Therefore:
$\lim\limits_{x \to a}f(x) = L~~$ if and only if $~~\lim\limits_{x \to a^-}f(x) = L = \lim\limits_{x \to a^+}f(x)$
