Calculus help

Question

Determine the following limit by either removing the “offending” factor or using
L’Hopital’s Rule:

 

 

lim x²+3x-28

^x→4 x²-x-12

Answer 1

The limit equals 11/7. We have x^2 + 3x - 28 = (x + 7)(x - 4) and x^2 - x - 12 = (x + 3)(x - 4), so 

 

(x^2 + 3x - 28)/(x^2 - x - 12) = (x + 7)/(x + 3)

 

for all x ≠ 4. Therefore

 

lim(x -> 4) (x^2 + 3x - 28)/(x^2 - x - 12) = lim(x -> 4) (x + 7)/(x + 3) = (4 + 7)/(4 + 3) = 11/7