Limit with cubic factor

**jacs** · April 5th 2011, 04:10 AM

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

**jacs** · April 5th 2011, 04:15 AM

Find $\lim_{x\to0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}$

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

although i am not sure how to do it

thanks

**FernandoRevilla** · April 5th 2011, 04:19 AM

Now, $a=2+x,\;b=2$ .

**amul28** · April 5th 2011, 04:28 AM

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

which is one of generalised form

$lim_{x\to 0}\frac{x^n-a^n}{x-a} = na^{n-1}$

**TheChaz** · April 5th 2011, 04:43 AM

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.

**TheEmptySet** · April 5th 2011, 06:03 AM

Originally Posted by jacs

Find $\lim_{x\to0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}$

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

although i am not sure how to do it

thanks

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

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

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

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

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