What is the Method for Factoring Polynomials with Limits?

[susan__t]

the question is:

The limit as x approaches -2 when f(x)= x+2/x³+8

I cannot factor it, or use a limit law (to my knowledge) and am simply having trouble finding the answer which is suppose to be 1/12.

Any help getting me started would be greatly appreciated

Susan

 

[morphism]

Can you factor a^3 - b^3?

 

[statdad]

Actually, you need to factor

<br /> a^3 + b^3<br />

instead of the difference of two cubes (assuming your function is

<br /> f(x) = \frac{x+2}{x^3 + 8}<br />
)

 

[morphism]

It's the same thing, because a^3 + b^3 = a^3 - (-b)^3.

The reason I chose to use difference instead of sum is because difference is usually more familiar, and might spark the right idea.

 

[statdad]

Certainly true and obvious when you know the steps that are to be done. I've been teaching mathematics and statistics long enough to realize that occasionally a student may not make that connection - not because of lack of ability, but because of frustration, nerves, haste, or many other reasons. In a situation like this, a little more direct approach can't hurt.
My previous post was not intended to be smarmy - I hope that no offense was taken, because none was meant.

 

[susan__t]

thank you so much both of you, I really had no idea what I could branch out of the equation and I would never have found that on my own

 

[statdad]

You are welcome Susan_t. Good luck with your studies.

 

[HallsofIvy]

WHENEVER a polynomial is 0 at x= a, then it has a factor of (x-a). Since x³+ 1 is 0 at x= -1, it MUST have (x+1) as a factor. Divide x³+ 1 by x+ 1 to find the other factor (which is, of course, x²+ x+ 1).

Notice that says that any time you have a ratio of polynomial and both numerator and denominator are 0 at x= a, each MUST have a factor of x-a which you can then cancel.