20. A rational function [tex]\frac{p(x)}{q(x)}[/tex] (where [tex]p,q[/tex] are polynomials in [tex]x[/tex]) has a removable discontinuity at [tex]x=a[/tex] if for some positive integer [tex]n[/tex] we can factorize
[tex]p(x) = (x-a)^n p^*(x)[/tex]
[tex]q(x) = (x-a)^n q^*(x)[/tex]
[tex]\implies \dfrac{p(x)}{q(x)} = \dfrac{(x-a)^n p^*(x)}{(x-a)^n q^*(x)} = \dfrac{p^*(x)}{q^*(x)}[/tex]
That is, we "remove" the discontinuity at [tex]x=a[/tex] because while [tex]x\neq a[/tex], we have [tex]\frac{x-a}{x-a}=1[/tex]. The discontinuity is still there, since [tex]\frac{p(a)}{q(a)}[/tex] is undefined, but in the limit sense we can essentially ignore the factors of [tex]x-a[/tex]. The graph of [tex]\frac{p(x)}{q(x)}[/tex] will have all the features of [tex]\frac{p^*(x)}{q^*(x)}[/tex] aside from a hole at [tex]x=a[/tex].
In this case, we have
[tex]f(x) = \dfrac{x+4}{x^2-x-20} = \dfrac{x+4}{(x+4)(x-5)}[/tex]
and when [tex]x\neq-4[/tex], we can cancel the factor of [tex]x+4[/tex] so that
[tex]f(x) = \begin{cases}\dfrac1{x-5} & \text{if }x \neq -4 \\\\ \text{DNE} & \text{if }x = -4\end{cases}[/tex]
("DNE" = "does not exist", i.e. is undefined. For some reason "undefined" wouldn't render...)
This means [tex]x=-4[/tex] is a removable discontinuity.
We cannot do the same with the factor of [tex]x-5[/tex], so in contrast [tex]x=5[/tex] is a non-removable discontinuity.
21. The pieces of [tex]f(x)[/tex] defined on [tex]x<2[/tex] and [tex]x>2[/tex] are themselves continuous since they are polynomials. Then the continuity of [tex]f(x)[/tex] over the entire real line depends on the point where the pieces meet.
Here we have
[tex]f(x) = \begin{cases}x+3 & \text{if }x\le2 \\ cx+6 & \text{if }x>2\end{cases}[/tex]
so the pieces meet at [tex]x=2[/tex]. Continuity at this point requires that the both limits from either side of [tex]x=2[/tex] be the same. This means
[tex]\displaystyle \lim_{x\to2^-} f(x) = \lim_{x\to2} (x+3) = 2+3 = 5[/tex]
[tex]\displaystyle \lim_{x\to2^+} f(x) = \lim_{x\to2} (cx+6) = 2c + 6[/tex]
Solve for [tex]c[/tex].
[tex]2c + 6 = 5 \implies 2c = -1 \implies \boxed{c = -\dfrac12}[/tex]
