Find $\displaystyle \lim_{x\rightarrow0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}$

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- Apr 5th 2011, 04:10 AMjacsLimit with cubic factor
Find $\displaystyle \lim_{x\rightarrow0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}$

- Apr 5th 2011, 04:15 AMjacsLimit with cubic factor
Find $\displaystyle \lim_{x\to0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}$

I suspect somehow i need to utlise $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

although i am not sure how to do it

thanks - Apr 5th 2011, 04:19 AMFernandoRevilla
Now, $\displaystyle a=2+x,\;b=2$ .

- Apr 5th 2011, 04:28 AMamul28
$\displaystyle lim_{x\to 0}\frac{{(2+x)^{\frac{1}{3}}}-2^{\frac{1}{3}}}{(2+x)-2}$

which is one of generalised form

$\displaystyle lim_{x\to 0}\frac{x^n-a^n}{x-a} = na^{n-1}$ - Apr 5th 2011, 04:43 AMTheChaz
As has been noted above, this is the "alternate" definition of the derivative (or "derivative at a point")..

If f(t) = t^(1/3)

f'(t) is the above limit, with t in place of 2.

So the above limit is f'(2) = (1/3)*2^(1/3-1)

OR.

View the numerator as one factor in the difference of cubes, and multiply by ...(something I'd rather not type) to "rationalize".

Then you can use direct substitution. - Apr 5th 2011, 06:03 AMTheEmptySet
This may be way off bases but I have never seen limits taught in a pre-algebra glass!

If you know about the derivative this this limit is

$\displaystyle \displaystyle \lim_{x \to 0}\frac{(2+x)^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}=f'(2)$

Where $\displaystyle f(x)=x^{\frac{1}{3}}$.

So as I said if you know about derivatives just take the derviative of f and plug in 2!