Thread: Limit with cubic factor

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

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

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

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

which is one of generalised form

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

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.

Originally Posted by jacs

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

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!