# LIMITS, not able to factor

1. Sep 16, 2008

### 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

2. Sep 16, 2008

### morphism

Can you factor a^3 - b^3?

3. Sep 16, 2008

Actually, you need to factor

$$a^3 + b^3$$

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

$$f(x) = \frac{x+2}{x^3 + 8}$$
)

4. Sep 16, 2008

### 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.

5. Sep 16, 2008

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.

6. Sep 16, 2008

### 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

7. Sep 17, 2008

You are welcome Susan_t. Good luck with your studies.

8. Sep 17, 2008

### HallsofIvy

Staff Emeritus
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.