Is it correct to say that the function $f(x)$ is everywhere continuous even if left side limit doesn't match with right side limit?

Question

If a function $$f(x)=\begin{cases} \frac{|x-4|}{x-4} & \textrm{ , } x\neq 4 \\ 0 & \textrm{ , } x=4 \end{cases}$$. Prove that $f$ is everywhere continuous except at $x=4$.

Rewrite $$f(x)=\begin{cases} -1 & \textrm{ , } x< 4 \\ 0 & \textrm{ , } x=4\\1 & \textrm{ , } x>4 \end{cases}$$

Now, $$\lim_{x\to4^-}f(x)\neq \lim_{x\to4^+}f(x)\neq f(4)$$, therefore $f(x)$ is not continuous at $x=4$.

Again, rewrite, $$f(x)=\begin{cases} -1 & \textrm{ , } x=3,2,1,0,-1,-2,... \\ 0 & \textrm{ , } x=4\\1 & \textrm{ , } x=5,6,7,8,9,10,.... \end{cases}$$ (interger values of $x$ are taken for convenience to understand)

I want to understand why (or how) $f(x)$ is everywhere continuous excluding $x=4$.Is it correct to say that $f(x)$ is evrywhere continuous even if $\text {L.H. Limit} \neq \text {R.H. Limit}$?

Any help or explanation is greatly appreciated.

Answer 1

Take any $x_0\ne4$. Then $f$ is constant in a full (maybe small) two-sided neighborhood of $x_0$, hence continuous at $x_0$. At $x_0=4$ the lefthand and the righthand limit of $f$ are unequal, hence $f$ is not continuous at $x_0=4$ (and cannot be made continuous there by a clever redefinition of $f(4)$).

It is therefore absolutely correct to say that $f$ is everywhere on ${\mathbb R}$ continuous, except at $x_0=4$.

Answer 2

This just my reply to your comment, but it is too large for the comment.

To "really see" that the limits are different, rewrite the function as $$f(x)=\begin{cases} \frac{-x+4}{x-4} & \textrm{ , } x< 4 \\ 0 & \textrm{ , } x=4\\\frac{x-4}{x-4} & \textrm{ , } x>4 \end{cases}.$$

Take limits and observe that they are different.