# Thread: find a such that and the limit

1. ## find a such that and the limit

Is there a number $\displaystyle \alpha$ such that

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

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

I tried first to factor the numerator into

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

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

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

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

3. so then

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

4. I like to see the limit signs

Originally Posted by bigwave
so then

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