Q.Discuss the continuity of the following function at the point(s) or on the interval indicated against it: $f(x) = \dfrac{x^2-3x-10}{x-5}$, for $3 \le x \le 6$, $x \ne 5$, $= 10$, for $x=5$, $= \dfrac{x^2-3x-10}{x-5}$, for $6 < x \le 9$.

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 cancell …