Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit

Question

Problem

If there exists $a \in \mathbb{R}$ such that:

$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$

has limit in $-2$. If such $a$ exists what is limit in $-2$ ?

Attempt to solve

My idea was first to try factorize denominator and then find factor of nominator in a way that these two cancel each other.

factorizing denominator gives:

$$ \lim_{x \rightarrow -2}\frac{3x^2+ax+a+3}{(x-1)(x+2)} $$

Now if it is possible to find solution to $3x^2+ax+a+3=0$ when $x=-2$ $$ 3(-2)^2+a(-2)+(-2)+3=0 $$ $$ 12-2a-2+3=0 $$ $$ 2a=13 \iff a = \frac{13}{2} $$

factorizing nominator gives:

$$ 3x^2+\frac{13x}{2}+\frac{19}{2}=0 $$ $$ x= \frac{-\frac{13}{2}\pm \sqrt{(\frac{13}{2})^2-4\cdot 3 \cdot (\frac{19}{2})}}{2\cdot 3} $$

Only problem is this equation is never zero with $x\in \mathbb{R}$ since there is negative value under square root. Now there is contradiction between $a=\frac{13}{2}$ and that equation $3x^2+ax+a+3=0$ when $a=\frac{13}{2}$ and $x=-2$

$$ 3(-2)^2+\frac{13}{2}\cdot (-2)+\frac{13}{2}+3 \neq 0 $$

There is clearly something wrong but i cannot see where this went wrong.

Answer 1

Since the denominator at $x=-2$ is equal to $0$ it is necessary that also the numerator at $x=-2$ is equal to $0$ that is

$$3(-2)^2+a(-2)+a+3 =0 \implies a=15 $$

and therefore

$$\lim_{x \rightarrow -2}\frac{3x^2+15x+18}{x^2+x+2}=\lim_{x \rightarrow -2}\frac{3(x+2)(x+3))}{(x-1)(x+2)}$$

Answer 2

When you substituted $x=-2$ in $3x^2+ax+a+3$ you incorrectly put $(-2)$ in place of some of the $a$-s. The remaining part appears to be the follow-up of said mistake of algebra.

Answer 3

You have $3x^2+ax+a+3=0$ and you set $x = -2$ and got $3(-2)^2+a(-2)+(-2)+3=0$

But you should have gotten $3(-2)^2+a(-2)+a+3=0$ .

That is an "$a$" between the $ax$ and the $3$. It is not an "$x$".

So you should have gotten $12 -2a + a + 3 = 0$ or $15 -a = 0$ or $a = 15$

.......

so $3x^2 + 15x+18 = 0$ or in other words $3(x^2 + 5x + 6)=0$ so $3(x+2)(x+3)=0$ so you get $\lim \frac {3(x+2)(x+3)}{(x-1)(x+2)} = \lim \frac {3(x+3)}{x-1} = \frac {3(-2 +3)}{-2-1}=-1$.