1. Feb 23, 2012

mtayab1994

1. The problem statement, all variables and given/known data
find the limit of:

$$\lim_{x\rightarrow2}\frac{1}{x-2}-\frac{12}{x^{3}-8}$$

3. The attempt at a solution

Any help guys? Should i cross multiply then divide and see what i get or is there something else.(besides l'hospital's rule and taylor series because i haven't learned them yet.)

2. Feb 23, 2012

HallsofIvy

Staff Emeritus
I can't imagine why you would think "cross multiply" here. Of course, what you should do is the subtraction indicated in the problem"
$$\frac{1}{x- 2}- \frac{12}{x^3- 8}= \frac{?}{x^3- 8}$$

3. Feb 23, 2012

Staff: Mentor

Cross multiply? You do that when you have an equation that involves two fractions.

Combine the two fractions, and then go from there. Note that the difference of cubes can be factored: a3 - b3 = (a - b)(a2 + ab + b2).
L'Hopital's Rule cannot be applied to this problem because what you have is a difference, not a quotient.

4. Feb 23, 2012

mtayab1994

ok i got:

$$\lim_{x\rightarrow2}\frac{x+4}{x^{2}+2x+4}=\frac{1}{2}$$

is that correct?

5. Feb 23, 2012

Staff: Mentor

Yes - perfect!

6. Feb 23, 2012

Thanx.