Rigorous proof of continuity at a if and only if left and right limits equal f(a)

Question

My question is the following. I use a lot of times that a function is continuous at $a$ if and only if

$$\lim_{x\to a^{-}}f(x) = f(a) = \lim_{x\to a^{+}}f(x) $$

But I've never found a rigorous proof (that I can still understand!). Maybe with $\epsilon - \delta$ definitions? Can somebody help me? Thanks!

Answer 1

A map $f$ is continuous at $a$ iff for every $\varepsilon > 0$ there is some $\delta > 0$ such that $0 < |x-a| < \delta$ only if $|f(x) - f(a)| < \varepsilon$; but $0 < |x-a| < \delta$ iff $-\delta < x-a < 0$ and $0 < x-a < \delta$; hence $f$ is continuous at $a$ iff $\lim_{x \to a-}f(x) = f(a) = \lim_{x \to a+}f(x)$.

Answer 2

$\lim_{x\to a^{+}}f(x)=f(a)$ means that for any $\epsilon >0$ there exists $\delta_1>0$ such that $$x\in (a,a+\delta_1)\implies |f(x)-f(a)|<\epsilon.$$

$\lim_{x\to a^{-}}f(x)=f(a)$ means that for any $\epsilon >0$ there exists $\delta_2>0$ such that $$x\in (a-\delta_2,a)\implies |f(x)-f(a)|<\epsilon.$$

Thus, given $\epsilon>0$ there exists $\delta=\min\{\delta_1,\delta_2\}$ such that

$$|x-a|\le \delta \implies |f(x)-f(a)|<\epsilon,$$ that is, $f$ is continuous at $a.$