# Calculus

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Is there a number "a" such that the limit as x approaches -2, f(x) approaches (3x^2+ax+a+3)/(x^2+x-2) exist?

• Calculus -

As x → -2, the denominator approaches zero.
Fo the limit to exist, the numerator must also be zero, which means:
if f(x)=(3x²+ax+a+3)
the f(-2)=0
So solve for a in f(-2)=0, AND check using l'Hôpital's rule that the limit exists for the assumed value of a. You may need to apply l'Hôpital's rule more than once.

I have 3 as the limit.

• Calculus -

yes, if a=15

Since the limit exists, both numerator and denominator must have the factor x+2
by synthetic division
(3x^2+ax+a+3) รท (x+2) leaves no remainder, so a = 15
then the question becomes
lim (3x^2 + 15x + 18)/(x^2 + x - 2)
x ---> -2
= lim (3(x+2)(x+3)/((x+2)(x-1))
x ---> -2
= lim 3(x+3)/(x-1)
x ---> -2
= 3(1)/(-3) = -1

• Calculus -

Thanks Reiny, I took the derivative a second time without checking if both numerator and denominator are zero.

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