## Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

Limits, differentiation, related rates, integration, trig integrals, etc.
jsel
Posts: 7
Joined: Mon Oct 11, 2010 11:07 pm
Contact:

### Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

$\mbox{49. Is there a number }\, a\, \mbox{ such that}$

. . . . . . .$\lim{x\rightarrow -2}\, \frac{3x^2\, +\, ax\, +\, a\, +\, 3}{x^2\, +\, x\, -\, 2}$

$\mbox{exists? If so, find the value of }\, a\, \mbox{ and the value of the limit.}$

hello everyone, this problem has got me particularly confused. Can someone point me in the right direction?
Last edited by jsel on Tue Oct 12, 2010 1:50 am, edited 1 time in total.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

$\mbox{49. Is there a number }\, a\, \mbox{ such that}$

. . . . . . .$\lim{x\rightarrow -2}\, \frac{3x^2\, +\, ax\, +\, a\, +\, 3}{x^2\, +\, x\, -\, 2}$

$\mbox{exists? If so, find the value of }\, a\, \mbox{ and the value of the limit.}$

hello everyone, this problem has got me particularly confused. Can someone point me in the right direction?

Yes there is a value. What is the denominator going to? What then does the numerator have to go to in order for the limit to exist?

jsel
Posts: 7
Joined: Mon Oct 11, 2010 11:07 pm
Contact:

### Re: Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

I know the denominator will be (x+2)(x-1). For the numerator, I cannot think of a factorable quadratic that will end cancel out any of the bottom terms.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Since you have the factors of the denominator, and since you know that you need the numerator to have one of the same factors, you might want to try dividing the numerator by one of the denominator's factors.

See where this leads, keeping in mind that, for the factor to divide evenly into the numerator, the remainder, upon division, must equal zero.

jsel
Posts: 7
Joined: Mon Oct 11, 2010 11:07 pm
Contact:

### Re: Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

Ok, I see where you are going but how do I go about polynomial division with "a" in the quadratic?

jsel
Posts: 7
Joined: Mon Oct 11, 2010 11:07 pm
Contact:

### Re: Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

OK I found a way to find a. But it is very odd-ballish and inefficient I think.

I began doing synthetic division of the quadratic by the -2 root. I stopped half way through and wrote an equation to solve for a so that there would be no remainder. If anyone has a better way of finding "a" please share.

Sorry about the MS paint image... I would have hated typing that out.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
$\begin{array}{r|rrrrr}-2&3&\,&a&\,&a+3\\\,&\,&\,&-6&\,&12-2a\\\hline\\\,&3&\,&a-6&\,&15-a\end{array}$

$15\, -\, a\, =\, 0,\, \mbox{ so }\, 15\, =\, a$
I believe what you've done is very likely what was expected.

Note: You can check your result by plugging "15" in for "a", and verifying that the result simplifies as desired.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Find 'a' so lim,x->-2,(3x^2+ax+a+3)/(x^2+x-2) exists

or you could think of it this way...

the only way the limit will exist is if the numerator is zero when x=-2

so take $3x^2+ax+a+3$

replace x by -2

and solve the equation (equal to zero) for a

jsel
Posts: 7
Joined: Mon Oct 11, 2010 11:07 pm
Contact:

### Re:

$\begin{array}{r|rrrrr}-2&3&\,&a&\,&a+3\\\,&\,&\,&-6&\,&12-2a\\\hline\\\,&3&\,&a-6&\,&15-a\end{array}$

$15\, -\, a\, =\, 0,\, \mbox{ so }\, 15\, =\, a$
I believe what you've done is very likely what was expected.

Note: You can check your result by plugging "15" in for "a", and verifying that the result simplifies as desired.
How did you format your text like that?

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Re:

$\begin{array}{r|rrrrr}-2&3&\,&a&\,&a+3\\\,&\,&\,&-6&\,&12-2a\\\hline\\\,&3&\,&a-6&\,&15-a\end{array}$

$15\, -\, a\, =\, 0,\, \mbox{ so }\, 15\, =\, a$
I believe what you've done is very likely what was expected.

Note: You can check your result by plugging "15" in for "a", and verifying that the result simplifies as desired.
How did you format your text like that?
if you just 'quote' the text stapel_eliz wrote you would see that it is...

Code: Select all

$$\begin{array}{r|rrrrr}-2&3&\,&a&\,&a+3\\\,&\,&\,&-6&\,&12-2a\\\hline\\\,&3&\,&a-6&\,&15-a\end{array}$$ $$15\, -\, a\, =\, 0,\, \mbox{ so }\, 15\, =\, a$$