# Homework Help: Another Limits Problem

1. Sep 4, 2009

### efekwulsemmay

1. The problem statement, all variables and given/known data
I have to find the tangent to the curve the given point $$\left(-2,-8\right)$$ for the equation $$y=x^{3}$$.

2. Relevant equations

$$f(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$$

3. The attempt at a solution
I started the normal way of substitution:

$$\lim_{h\rightarrow0} \frac{f(-2+h)-f(-2)}{h}$$

which goes to:

$$\lim_{h\rightarrow0} \frac{(-2+h)^{3}-(-2)^{3}}{h}$$

This is where I get stuck. I know that you factor cubic functions by:
$$(a+b)(a^{2}-ab+b^{2})$$

However, when I do this and multiply it out I eventually end up with:

$$\lim_{h\rightarrow0} \frac{h^{3}}{h}$$

My solutions manual says I should be getting:

$$\lim_{h\rightarrow0} \frac{-8+12h-6h^{2}+h^{3}+8}{h}$$

I don't understand how it got to this point and it doesn't say. Help me please?

2. Sep 4, 2009

### VietDao29

Well, depending on what your a, and b are, you may have used the wrong formula. There are 2 formulae to factor cubes:

1. a3 + b3 = (a + b)(a2 - ab + b2) (which is what you have written, also called, Sum of 2 Cubes Formula).
2. a3 - b3 = (a - b)(a2 + ab + b2) (and this is called Difference of 2 Cubes Formula).

Your book is correct. So, you should re-check the steps again. :( Or, if you cannot find any mistakes in your work, you can always post it here, and we'll be more than willing to help you spot out the errors.

3. Sep 4, 2009

### sara_87

always factor out cubic functions like this:
(a+b)3= a3 + 3a2b + 3ab2 +b3
for the numerator:

(-2+h)3 - (-2)3
= (-2)3 + 3(-2)2(h) + 3(-2)(h2) + h3 - (-8)
= -8+12h-6h2+h3+h

4. Sep 4, 2009

### VietDao29

Giving out complete solution is indeed against the forums' rule!!!! :grumpy: You should give the OP some chances to think, and work by himself, instead of feeding him the answer like that. Feeding complete answer is definitely not our goal here. :(

5. Sep 4, 2009

### sara_87

woops, i think i gave more help than i should have
i wish someone would 'accidently' do that to one of my questions :shy:

6. Sep 4, 2009

### sara_87

I was trying to help.

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