Math  /  Calculus

QuestionIs there a number a such that the following limit exists? (If an answer does not exist, enter DNE.) limx23x2+ax+a+9x2+x2\lim _{x \rightarrow-2} \frac{3 x^{2}+a x+a+9}{x^{2}+x-2}
Find the value aa. a=7a=-7
Evaluate the limit. 55

Studdy Solution

STEP 1

1. The limit involves a rational function, which may have indeterminate forms at x=2 x = -2 .
2. To find the limit, we need to ensure the expression is not undefined at x=2 x = -2 .
3. We will factor the denominator and possibly the numerator to simplify the expression.

STEP 2

1. Factor the denominator.
2. Determine conditions for the numerator to have a common factor with the denominator.
3. Simplify the expression.
4. Evaluate the limit.

STEP 3

Factor the denominator x2+x2 x^2 + x - 2 .
x2+x2=(x+2)(x1) x^2 + x - 2 = (x + 2)(x - 1)

STEP 4

To avoid an indeterminate form at x=2 x = -2 , the numerator 3x2+ax+a+9 3x^2 + ax + a + 9 must also be zero at x=2 x = -2 .
Substitute x=2 x = -2 into the numerator:
3(2)2+a(2)+a+9=0 3(-2)^2 + a(-2) + a + 9 = 0 122a+a+9=0 12 - 2a + a + 9 = 0 21a=0 21 - a = 0
Solve for a a :
a=21 a = 21

STEP 5

The problem states a=7 a = -7 , so we need to check if this value satisfies the condition.
Substitute a=7 a = -7 into the numerator and check if it is zero at x=2 x = -2 :
3(2)27(2)7+9=0 3(-2)^2 - 7(-2) - 7 + 9 = 0 12+147+9=0 12 + 14 - 7 + 9 = 0 28=0 28 = 0
This does not equal zero, so a=7 a = -7 does not satisfy the condition. The correct value of a a should be 21.

STEP 6

With a=21 a = 21 , the numerator becomes 3x2+21x+21+9 3x^2 + 21x + 21 + 9 .
Factor the numerator:
3x2+21x+30=3(x2+7x+10) 3x^2 + 21x + 30 = 3(x^2 + 7x + 10) =3(x+2)(x+5) = 3(x + 2)(x + 5)

STEP 7

Simplify the expression by canceling the common factor (x+2) (x + 2) :
3(x+2)(x+5)(x+2)(x1)=3(x+5)x1 \frac{3(x + 2)(x + 5)}{(x + 2)(x - 1)} = \frac{3(x + 5)}{x - 1}
Evaluate the limit as x2 x \rightarrow -2 :
limx23(x+5)x1=3(2+5)21 \lim_{x \rightarrow -2} \frac{3(x + 5)}{x - 1} = \frac{3(-2 + 5)}{-2 - 1} =3×33=3 = \frac{3 \times 3}{-3} = -3
The limit does not exist for a=7 a = -7 . The correct value of a a should be 21, and the evaluated limit is 3 \boxed{-3} .

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