Answer:
Step-by-step explanation:
Limits exist for:
[tex]\lim_{x \to4 \4} \frac{x^2 -x-12}{4-x}[/tex] = [tex]\frac{(x-4)(x+3)}{4-x}[/tex] = ( - 1 )( x + 3 ) = - 7
[tex]\lim_{x \to \ 2} \frac{x^2 +2x-8}{2-x}[/tex] = [tex]\frac{(x+4)(x-2)}{2-x}[/tex] = ( - 1 )( x + 4 ) = - 6
[tex]\lim_{x \to2 } \frac{x^2 +3x - 10}{x^2 -4}[/tex] = [tex]\frac{(x+5)(x-2)}{(x+2)(x-2)}[/tex] = [tex]\frac{x+5}{x+2}[/tex] = [tex]\frac{7}{4}[/tex]
