Does anyone knows how to show:
limit of f(x), x approaches a, equals L
if and only if
limit of f(x), x approaches a from the right = to limit of f(x), x approaches a from the left = L
Using the precise definition of limit.
Let $\displaystyle f$ be defined on an open interval $\displaystyle I$ containing $\displaystyle a$ except possibly at $\displaystyle a$. With domain $\displaystyle D$.
Let, $\displaystyle \lim_{x\to a}f(x) = L$.
That means,
$\displaystyle 0 < |x-a| < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
We want to show,
$\displaystyle \lim_{x\to a^+}f(x) = L \mbox{ and }\lim_{x\to a^-}f(x) = L$.
Meaning,
$\displaystyle 0 < a - x < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
$\displaystyle 0 < x - a < \delta \implies x\in D \mbox{ and }|f(x)-L|<\epsilon$.
Do you see? Note that $\displaystyle |x-a|$ can either be $\displaystyle x-a \mbox{ or }a-x$ depending whether $\displaystyle x$ is larger than $\displaystyle a$ or not. But the point is that bpth of these quantieis $\displaystyle x-a, \ a-x$ are containted in the meaning of $\displaystyle 0<|x-a|< \delta$.