# Is there a number "a" such that the equation below exists? If so what is the value of "a" and its limit.

## ${\lim}_{x \to - 2} \frac{3 {x}^{2} + a x + a + 3}{{x}^{2} + x - 2}$

Oct 2, 2016

$a = 15$

${\lim}_{x \to - 2} \frac{3 {x}^{2} + 15 x + 18}{{x}^{2} + x - 2} = - 1$

#### Explanation:

${x}^{2} + x - 2 = \left(x + 2\right) \left(x - 1\right)$

So the denominator contains exactly one factor $\left(x + 2\right)$

So in order that $\frac{3 {x}^{2} + a x + a + 3}{{x}^{2} + x - 2}$ has a limit as $x \to - 2$, the only requirement is that:

$3 {x}^{2} + a x + \left(a + 3\right) \text{ }$ is divisible by $\left(x + 2\right)$

Let $f \left(x\right) = 3 {x}^{2} + a x + \left(a + 3\right)$

This is divisible by $\left(x + 2\right)$ if and only if $f \left(- 2\right) = 0$

Substituting $x = - 2$ we have:

$f \left(- 2\right) = 3 {\left(\textcolor{b l u e}{- 2}\right)}^{2} + a \left(\textcolor{b l u e}{- 2}\right) + a + 3$

$\textcolor{w h i t e}{f \left(- 2\right)} = 12 - 2 a + a + 3$

$\textcolor{w h i t e}{f \left(- 2\right)} = 15 - a$

So we require $a = 15$

With this value of $a$:

$f \left(x\right) = 3 {x}^{2} + 15 x + 18 = 3 \left({x}^{2} + 5 x + 6\right) = 3 \left(x + 2\right) \left(x + 3\right)$

$\frac{3 {x}^{2} + 15 x + 18}{{x}^{2} + x - 2} = \frac{3 \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 2}}}\right) \left(x + 3\right)}{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 2}}}\right) \left(x - 1\right)} = \frac{3 \left(x + 3\right)}{x - 1}$

So:

${\lim}_{x \to - 2} \frac{3 {x}^{2} + 15 x + 18}{{x}^{2} + x - 2} = {\lim}_{x \to - 2} \frac{3 \left(x + 3\right)}{x - 1} = \frac{3 \left(\textcolor{b l u e}{- 2} + 3\right)}{\textcolor{b l u e}{- 2} - 1} = \frac{3}{- 3} = - 1$