A bit late, but better than never.
24. For [tex]x\neq1[/tex], we have
[tex]\dfrac{9x^2 - x - 8}{x - 1} = \dfrac{(x-1)(9x+8)}{x-1} = 9x+8[/tex]
Then as [tex]x[/tex] approaches 1, we have
[tex]\displaystyle \lim_{x\to1} f(x) = \lim_{x\to1} (9x+8) = 17 \neq 0[/tex]
so the function is not continuous at [tex]x=1[/tex]. It is thus continuous on the interval union [tex](-\infty,0)\cup(0,\infty)[/tex]. You can also write this as "[tex]x<0[/tex] or [tex]x>0[/tex]".
25. When [tex]x[/tex] approaches 2 from the left (when [tex]x<2[/tex]), we have
[tex]\displaystyle \lim_{x\to2^-} f(x) = \lim_{x\to2} (5-x) = 3[/tex]
When [tex]x[/tex] approaches 2 from the right [tex](x>2)[/tex], we have
[tex]\displaystyle \lim_{x\to2^+} f(x) = \lim_{x\to2} (2x-3) = 1[/tex]
so the function is not continuous at [tex]x=2[/tex]. Thus it's continuous on [tex](-\infty,2)\cup(2,\infty)[/tex], or "[tex]x<2[/tex] or [tex]x>2[/tex]".
