Find $a$ if $\lim_{x\to-2}⁡\frac{3x^2+ax+a+3}{x^2+x-2}$ exists.

Question

Is there a 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 $a$

I desperately need help in trying to work this out. I already tried out factorising the numerator to get an expression of $(x+2)$ to cancel the denominator's $x+2$ but so far no luck.

Answer 1

Let $p(x)=3x^2+ax+a+3$ and $q(x)=x^2+x-2=(x-1)(x+2).$ Then we have:

$$p(-2)=15-a.$$

Case 1: $a \ne 15.$ Then $ \lim_{x \to -2}p(x) \ne 0,$ and $ \lim_{x \to -2} q(x)=0.$

Hence the limit in question does not exist.

Case 2: $a=15$. Then $ \frac{p(x)}{q(x)}= \frac{3(x+1)}{(x-1)} \to -1$ as $x \to -2.$

Answer 2

Ok so we have $x^2+x-2=(x+2)(x-1)$ and $-2+2=0$ so we need the numerator to have an $x+2$ in its factorization. Call the numerator $P(x)$, now if theres an $x+2$ in its factorization then we know $P(x)=R(x)(x+2)$ where $R(x)$ is a degree 1 polynomial(this is by the polynomial dividing algorithm), which means $P(-2)=0$ so we have $P(-2)=12-2a+a+3=15-a=0$ so $a=15$. We have $3x^2+15x+18=3(x+3)(x+2)$. So the limit is $\frac 3 {-3}=-1$.

Answer 3

We want $x=-2$ to be a root of $3x^2+ax+a+3$, hence $a=15$.