# Thread: Two limits of indeterminant form.

1. ## Two limits of indeterminant form.

how do you solve this two limits?

it's kinda urgent.

thanks alot!

2. Originally Posted by kiklop
how do you solve this two limits?

it's kinda urgent.

thanks alot!
For the first one, you might try applying the generalised binomial theorem, simplifying and then taking the limit.

For the second, several approaches are possible. ONe approach would be to use l'Hopital's Rule. See limit of &#40;1 - Cos&#91;2x&#93;&#41;&#47;&#40;x Tan&#91;3x&#93;&#41; as x approaches 0 - Wolfram|Alpha and click on Show steps.

If you need more help, please show all your working and say where you get stuck.

3. No need for Binomial Thorem...

$\displaystyle x\left(x - \sqrt[3]{x^3 + 4x}\right) = x\left[x - \sqrt[3]{x^3\left(1 + \frac{4}{x^2}\right)}\right]$

$\displaystyle = x\left[x - x\sqrt[3]{1 + \frac{4}{x^2}}\right]$

$\displaystyle = x^2\left(1 - \sqrt[3]{1 + \frac{4}{x^2}}\right)$.

$\displaystyle = \frac{1 - \sqrt[3]{1 + \frac{4}{x^2}}}{x^{-2}}$.

This now goes to $\displaystyle \frac{0}{0}$, so L'Hospital's Rule can be used.