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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:**344**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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jsel wrote:

. . . . . . .

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?

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:**1687**Joined:**Mon Dec 08, 2008 4:22 pm-
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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.

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

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

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.

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
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- Martingale
**Posts:**344**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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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

replace x by -2

and solve the equation (equal to zero) for a

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

so take

replace x by -2

and solve the equation (equal to zero) for a

stapel_eliz wrote:

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:**344**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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jsel wrote:stapel_eliz wrote:

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

Code: Select all

`[tex]\begin{array}{r|rrrrr}-2&3&\,&a&\,&a+3\\\,&\,&\,&-6&\,&12-2a\\\hline\\\,&3&\,&a-6&\,&15-a\end{array}[/tex]`

[tex]15\, -\, a\, =\, 0,\, \mbox{ so }\, 15\, =\, a[/tex]