Hello everyone, I would like to ask a question.

Knowing that:

$\displaystyle sinx = x - \frac{x^3}{6} +o(x^3)$

How do I find..:

$\displaystyle sin^3 \sqrt(x)$

$\displaystyle sin \sqrt(x) = \sqrt(x) - \frac{\sqrt(x^3)}{6}$

Woops... I just realized how it should be done, I just had to do that series cubed.

i can't delete the post, sorry for bothering guys. >.<'