Q.Examine the continuity of $f(x) = \dfrac{x^2-9}{x-3}$, for $x \ne 3$, $= 8$ for $x = 3$.

A removable discontinuity is the mildest kind of discontinuity a function can have at a point $x=a$: the two-sided limit $\lim_{x\to a} f(x)$ genuinely exists as a finite number, but the function fails to be continuous there only because either $f(a)$ was never defined at all, or it was assigned some other value that does not match that limit. Because the limit itself is perfectly well-behaved, the discontinuity can always be patched by simply defining (or redefining) $f(a)$ to equal $\lim_{x\to a} f(x)$; the repaired function is then continuous at $a$, and if the original function had no value there to begin with, this patched version is called its extension. The most common source of this situation is a rational expression where a factor in the numerator exactly cancels a matching factor in the denominator — the cancelling factor is what caused the original $0/0$ form, and once it is cancelled algebraically, evaluating the simplified expression at $a$ gives the value the extension should use. The same idea also shows up whenever a function is deliberately given the "wrong" constant at a single point: the fix is always the same, replace that constant with the true limit.