# Thread: find a such that and the limit

1. ## find a such that and the limit

Is there a number $\alpha$ such that

$
\lim_{x \to -2}\frac{3x^2+\alpha x + \alpha + 3}{x^2 + x - 2}
$

exists? If so find the value of
$\alpha$ and the value of the Limit

I tried first to factor the numerator into

$\left(x-1\right)\left(x+2\right)$

thinking this would reveal how to factor the numerator
to find
$\alpha$ but nothing became obvious.

$\alpha = 15$ limit $= -1$

2. you need the numerator to be 0 when x=-2
Inserting -2 for x does make $\alpha=15$
THEN factor the numerator and cancel the x+2 terms.

3. so then

$\lim_ {x \to -2}
\frac
{3\left(x+2\right)\left(x+3\right)}
{\left(x+2\right)\left(x -1\right)}
\Rightarrow
\frac
{3\left(x+3\right)}
{\left(x -1\right)}
\Rightarrow
\frac
{3\left(-2+3\right)}
{\left(-2 -1\right)}
\Rightarrow
\frac
{3\left(1\right)}
{\left(-3\right)}
\Rightarrow
-1
$

4. I like to see the limit signs

Originally Posted by bigwave
so then

$\lim_ {x \to -2}
\frac
{3\left(x+2\right)\left(x+3\right)}
{\left(x+2\right)\left(x -1\right)}
= \lim_ {x \to -2}
\frac
{3\left(x+3\right)}
{\left(x -1\right)}
= \lim_ {x \to -2}
\frac
{3\left(-2+3\right)}
{\left(-2 -1\right)}
=
\frac
{3\left(1\right)}
{\left(-3\right)}
=
-1
$