Finding the value of $a$ of an undefined limit

Question

Is there a real number a such that $$\lim_{x\to -2} \frac {3x^2 + ax + a + 3} {x^2 + x − 2}$$ exists? If so, find the value of a.

Here are my thought process:

I see that the denominator equals 0, and I have two conclusions:

1) If $3x^2 + ax + a + 3 = 0$, factorisation can be done to find the limit of the function.

2) If $3x^2 + ax + a + 3 ≠ 0$, the limit tends to infinity (which I don't think is very much applicable in this question.

The problem is I don't know how do i derive the values of $a$ with the given information. Thanks.

Answer 1

We need $$3(-2)^2+(-2)a+a+3=0,$$ otherwise the limit does not exist.

Thus, $a=15$ and $$\lim_{x\to -2} \frac {3x^2 + ax + a + 3} {x^2 + x − 2}=\lim_{x\rightarrow-2}\frac{3(x+2)(x+3)}{(x+2)(x-1)}=-1.$$

Answer 2

Limit is finite implies limit exists hence numerator has to be $0$ now simply plug in $x=-2$ and conclude that $a=15$

Answer 3

hint: You need $3(-2)^2 + a(-2) + a + 3 = 0$ because you want $x+2$ to be a factor of the numerator so that it will cancel out with the same factor $x+2$ of the denominator....

Answer 4

You started fine. Now, if $x=-2$, then $3x^2+ax+a+3=15-a$. So, if $a\neq15$, there is no limit (in $\mathbb R$). But$$\lim_{x\to-2}\frac{3x^2+15x+18}{x^2+x-2}=\lim_{x\to-2}\frac{6x+15}{2x+1}=-1.$$