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September 20, 2014

September 20, 2014

Posted by **John** on Saturday, September 18, 2010 at 7:00pm.

- Calculus -
**MathMate**, Saturday, September 18, 2010 at 7:14pmAs 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 -
**Reiny**, Saturday, September 18, 2010 at 7:27pmyes, 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) ÷ (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 -
**MathMate**, Saturday, September 18, 2010 at 8:24pmThanks Reiny, I took the derivative a second time without checking if both numerator and denominator are zero.

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