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

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.

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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
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 $3x^{2}$: $\lim_{h\to0}\frac{(x+h)^{3}-x^{3}}{h}=\lim_{h\to0}\frac{h^{3}+3h^{2}x+3hx^{3}}{h}=\lim_{h\to0}h^{2}+3hx+3x^{2}=3x^2$.

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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. – user38268 Jan 15 '12 at 3:37
If I use wolfram on every single problem I don't think I will learn anything though. – user138246 Jan 15 '12 at 3:45

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

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That should be $3x^2+3 h x + h^2$, not $3x + 3 h + h^2$, no? – DSM Jan 15 '12 at 2:55

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.

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