Why are removable discontinuities even discontinuities at all? If I have, for example, the function
$$f(x)=\frac{x^2+x-6}{x-2}$$
there will be a removable discontinuity at $x=2$, yes?
Why does this discontinuity exist at all if the function can be simplified to $f(x)=x+3$?
I suppose the answer is that you can't simplify it because you can't divide by something that could potentially equal $0$.
But what if you start with $f(x)=x+3$? What's stopping you from multiplying it by $1$ in the form $\frac{x-2}{x-2}$? Can multiplying by $1$ really introduce a discontinuity to the function?


What's stopping you from multiplying [a continuous function] by $1$ in the form $\frac{x−2}{x−2}$? Can multiplying by $1$ really introduce a discontinuity to the function?

Ah, but $\frac{x−2}{x−2}$ is not equal to $1$.  Or, to be more, precise, it is only equal to $1$ for values of $x$ that are different from $2$.  If $x=2$, then $\frac{x−2}{x−2}$ is undefined.
And that's more or less the answer to your question:  If you take a given function $f(x)$ that is continuous on all of $\mathbb{R}$, and you multiply it by a function $g(x)$ that is equal to $1$ everywhere except at a single isolated point, at which point $g(x)$ is undefined, then the product $f(x)g(x)$ will be exactly the same as $f(x)$ everywhere except at that point, where it, too, will be undefined.
