Expanding Cube Roots: Solving Limits with Maclaurin Series

[snoopies622]

A quiz at the end of Steven Krantz's Calculus Demystified includes the following problem:
Find
$$\lim_{x \to \infty} [ \sqrt[3]{x+1}<br /> -<br /> \sqrt[3]{x} ]$$
I see how one can use the Maclaurin series to get
$$\sqrt[3]{x+1} = 1 + \frac {x}{3} - \frac {x^2}{9} + \frac {5 x^3}{81} + . . .$$
but trying it with the cube root of x gives me zero plus an endless series of undefined terms.
Is there a way to expand $\sqrt[3]{x}$ and solve this problem?

 

[zoki85]

But why do you want expand that in series?
Isn't it by simple high school algebra :

for x>0?

 

[snoopies622]

Oh yeah, you're right zoki85 - thanks!

I also just realized that I could simply define y = x + 1, plug in y-1 for all the x's in the series above and there's the answer to my other question. :)