Find the number a such that the limit exists. Then find the value of the limit

Question

lim x→−2

(3x^2 + ax + a +9)/ (x^2 +x -2)

Answer 1

This question uses the concept of 'solving limits using factoring'.

We start by factoring the denominator

x²+x-2 = (x+2)(x-1)

For our limit to exist, the numerator too should have a factor of (x+2) so that it cancels out from the denominator

This means, 3x²+ax+a+9 = (x+2)(3x+c) ------ Eqn 1

I take the other factor has 3x+c because the coefficient of x² is 3 and as our other factor is going to be (x+2), this factor should have 3x

Solving Eqn 1

3x²+ax+a+9 = 3x²+cx+6x+2c

Coefficients of x² are equal, lets equate the coefficient of x and constant on both sides of equation

c+6 = a

a+9 = 2c

Solving the above system of equation, we get

a=21

c=15

Thus, value of a is 21

The expression will become lim x-> -2 (3x²+21x+30 )/(x²+x-2). The answer for the limit will be -3

Answer 2

lim _x→−2(3x^2 + ax + a +9)/ (x^2 +x -2)

We can write (x^2 +x -2) as (x-1)(x+2). So, we need to find a value of a such that

3x^2 + ax + a +9 = (x+2)(something else). This means that x=-2 should be a zero of 3x^2 + ax + a +9, that is

3x^2 + ax + a +9=0 for x=-2

Plugging x=-2 we get

12-a+9=0, a=21.

Then

lim _x→−2(3x^2 + ax + a +9)/ (x^2 +x -2)

=lim _x→−2(3x^2 + 21x +30)/ (x^2 +x -2)

=3 lim _x→−2(x^2 + 7x + 10)/ (x^2 +x -2)

=3 lim _x→−2(x +2)(x+5) / ((x+2)(x -1))

=3 lim _x→−2(x+5) / (x -1) = 3*(-3) /(-3) =3