# Thread: limits, finding value of b

1. ## limits, finding value of b

Is there a number b such that $\lim_{x \to -1} \frac{2x^2+bx+3b}{x^2-x-2}$ exists? If so, find the value of b and the value of the limit.
I'm just going to plug -1 into the denominator and that gives 0.
$-1^2-(-1)-2=0$
So for the numerator I am going to plug in -1 as well and solve for b
$2(-1)^2+b(-1)+3b=0$
That gives me $b=-1$
And $\lim_{x \to -1} \frac{2x^2-x-3}{x^2-x-2}=\frac{5}{3}$

Am I doing it right? Thanks

2. Originally Posted by iamanoobatmath
I'm just going to plug -1 into the denominator and that gives 0.

$-1^2-(-1)-2=0$

So for the numerator I am going to plug in -1 as well and solve for b

$2(-1)^2+b(-1)+3b=0$

That gives me

$b=-1$

And

$\displaystyle\lim_{x \to -1} \frac{2x^2-x-3}{x^2-x-2}=\frac{5}{3}$

Am I doing it right? Thanks
If you want to find the limit as $x \to -1$

you must resolve your "divide-by-zero" problem.
Therefore, try to factor the denominator to see if we can deal with an $(x+1)$ factor,

as $(x+1)=0$ if $x=-1$

$\displaystyle\frac{2x^2+bx+3b}{x^2-x-2}=\frac{2x^2+bx+3b}{(x+1)(x-2)}$

Now, if we can factor the numerator, with $(x+1)$ as one of those factors,
we will be able to find the limit...

$2x^2+bx+3b=(x+1)(2x+c)$

$\Rightarrow\ 2x^2+bx+3b=2x^2+cx+2x+c=2x^2+(c+2)x+c$

Hence, comparing terms we have

$c+2=b,\;\;\;c=3b$

Hence

$3b+2=b\Rightarrow\ 2b=-2\Rightarrow\ b=-1$

You can then evaluate the limit if you like.

The limit is evaluated by simplifying the fraction

$\displaystyle\lim_{x \to -1}\frac{(x+1)(2x+c)}{(x+1)(x-2)}=\frac{2x+c}{x-2}$ evaluated at $x=-1$