Respuesta :
because ( (-2)³+8))/(-2+2)=0/0,
you use formula a³+b³°(a+b)(a²-ab+b²).
therefore (x³+8)/(x+2) = ( (x+2)(x²-2x+4))/(x+2) = x²-2x+4
so lim x->-2 (x²-2x+4) = (-2)²-2*(-2)+4=
=4+4+4=12.
Answer:
12
Step-by-step explanation:
note that x³ + 8 is a sum of cubes and factors as
x³ + 8 = (x + 2)(x² - 2x + 4), thus
lim x → - 2 [tex]\frac{x^3+8}{x+2}[/tex]
= lim x → - 2 [tex]\frac{(x+2)(x^2-2x+4)}{x+2}[/tex]
Cancel factor (x + 2) on numerator/denominator
= lim x → - 2 (x² - 2x + 4) ← evaluate by direct substitution
= (- 2)² - 2(- 2) + 4 = 4 + 4 + 4 = 12
