Is there a number "a" such that the equation below exists? If so what is the value of "a" and its limit.
#lim_(x->-2)(3x^2+ax+a+3)/(x^2+x-2)#
1 Answer
#a = 15#
#lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = -1#
Explanation:
#x^2+x-2 = (x+2)(x-1)#
So the denominator contains exactly one factor
So in order that
#3x^2+ax+(a+3)" "# is divisible by#(x+2)#
Let
This is divisible by
Substituting
#f(-2) = 3(color(blue)(-2))^2+a(color(blue)(-2))+a+3#
#color(white)(f(-2)) = 12-2a+a+3#
#color(white)(f(-2)) = 15 - a#
So we require
With this value of
#f(x) = 3x^2+15x+18 = 3(x^2+5x+6) = 3(x+2)(x+3)#
#(3x^2+15x+18)/(x^2+x-2) = (3(color(red)(cancel(color(black)(x+2))))(x+3))/((color(red)(cancel(color(black)(x+2))))(x-1)) = (3(x+3))/(x-1)#
So:
#lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = lim_(x->-2) (3(x+3))/(x-1) = (3(color(blue)(-2)+3))/(color(blue)(-2)-1) = 3/(-3) = -1#