Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\lim\limits_{h\to 0}\frac{(x+h)^3-x^3}{h}$$

I tried to factor the top using the difference of cubes formula I looked up online, that gave an incorrect answer. I also tried to use a conjugate making $(x+h)^3=a$ and then I plugged in x and h afterwards, that did not work at all but I did get an answer, just not the right one. I can easily factor out an h and cancel it out but it isn't the right answer.

share|improve this question
    
As an addendum to the answers below, since you mention using the difference of cubes factoring pattern $\left(a^3-b^3=(a-b)(a^2+ab+b^2)\right)$: $$(x+h)^3-x^3=(x+h-x)((x+h)^2+(x+h)x+x^2)$$ $$=h(x^2+2xh+h^2+x^2+xh+x^2)$$ $$=h(3x^2+3xh+h^2)$$ So, it should work. –  Isaac Jan 15 '12 at 4:14
add comment

3 Answers

up vote 8 down vote accepted

Observe that this is a special limit of the form: $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=x^{3}$. This limit (the derivative of $f(x)$) gives you the slope of $f(x)$ which in this case is $3x^{2}$.

If you have not yet learned about the derivative, then the following computation verifies that the limit is indeed $3x^{2}$: $\lim_{h\to0}\frac{(x+h)^{3}-x^{3}}{h}=\lim_{h\to0}\frac{h^{3}+3h^{2}x+3hx^{3}}{h}=\lim_{h\to0}h^{2}+3hx+3x^{2}=3x^2$.

share|improve this answer
    
Damn, I am incredibly bad at math. Somehow during the 45 minutes of doing that problem I never once did the math properly. –  user138246 Jan 15 '12 at 3:02
    
Well, I find that Wolfram Alpha is an excellent tool for finding limits when one is stuck. –  MathMajor Jan 15 '12 at 3:13
    
@MathMajor +1 Nice observation. –  fpqc Jan 15 '12 at 3:37
4  
If I use wolfram on every single problem I don't think I will learn anything though. –  user138246 Jan 15 '12 at 3:45
add comment

Actually you can simply expand$(x+h)^3$ and the term $x^3$ can be eliminated and the limit $\lim_{h\rightarrow0} 3x^2+3xh+h^2=3x^2$

share|improve this answer
    
That should be $3x^2+3 h x + h^2$, not $3x + 3 h + h^2$, no? –  DSM Jan 15 '12 at 2:55
add comment

I can see two ways to do this:

  • Expand the cube of the sum: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ and after that do routine algebra.

  • Factor the difference of two cubes. Generally $a^3-b^3 = (a-b)(a^2+ab + b^2)$. Apply that to the case where $a=x+h$ and $b=x$.

Although L'Hopital's rule could also be used, it relies on methods of differentiation whose validity is probably just what you're trying to prove.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.