prove that a function $f$ is continuous at $x_0$ iff $f(x_{0}^{-}) = f(x_{0}^{+}) = f(x_0)$

Question

prove that a function $f$ is continuous at $x_0$ iff $$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

By definition, a function $f$ is said to be continuous at $x=x_0$ if $\lim\limits_{x \rightarrow x_{0}} f(x) = f(x_0)$

It follows that for each $\epsilon>0$ there is a $\delta>0$ s.t. $$| f(x) - f(x_0)|< \epsilon$$ whenever $$|x-x_0|<\delta$$

Considering $$|x-x_0|<\delta$$ is true $$-\delta < x -x_0 < \delta$$ $$ x_0 - \delta < x < x_0 + \delta $$

Therefore, On the right of $x_0$, it follows that : $x_0 \leq x < x_0 + \delta$ such that $| f(x) - f(x_0)|< \epsilon$. It shows that the function f is continuous from the right at $x_0$, Therefore: $$\lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

Similarly, on the left of $x_0$, it follows that $x_0 - \delta < x \leq x_0$ such that $| f(x) - f(x_0)|< \epsilon$. It shows that the function f is continuous from the left at $x_0$. Therefore: $$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

We can conclude then that if $f$ is continuous at $x_0$, $f$ has to be continuous from both the left and the right at $x_0$,It follows that $$\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)$$

Is this correct? anything could have been done differently? Any input is much appreciated

Answer 1

Using the definition of the limit and find the definition of continuous functions

$\scriptstyle \lim \limits_{x\to x_0^-}f(x)=f(x_0)\iff \quad \forall \varepsilon_1>0,\,\exists \eta_1>0,\;\forall x\in \mathcal{D}f,\; \quad-\eta_1<x-x_0<0\implies |f(x)-f(x_0)|<\varepsilon_1 \qquad(1)$

$\scriptstyle \lim \limits_{x\to x_0^+}f(x)=f(x_0)\iff \quad \forall \varepsilon_2>0,\;\exists \eta_2>0,\;\forall x\in \mathcal{D}f,\; \quad 0<x-x_0<\eta_2\implies |f(x)-f(x_0)|<\varepsilon_2 \qquad (2)$

We choose $\varepsilon >0$, then with (1) and (2), $\eta_1$ and $\eta_2$ exist

$\delta =\min(\eta_1,\eta_2)$ then

$\forall \varepsilon>0,\;\exists \delta>0,\;\forall x\in \mathcal{D}f,\; \quad |x-x_0|<\delta \implies |f(x)-f(x_0|<\varepsilon$

which is the definition of continuous functions