Originally Posted by

**iMaths** oh right i forgot that. i keep think in limit you just ignore the denominator. so you differentiate the numerator and the denominator?

assuming you equation is: $\displaystyle \frac{\sqrt{2a^{3}x-x^4}-a\sqrt[3]{aax}}{a-\sqrt[4]{ax^3}}$

using L'hopital's rule you differentiate both top and bottom? Chain rule?

=$\displaystyle \frac{0.5(2a^{3}x-x^4)^{-1/2}(2a^3-4x^3)-(1/3)a^{5/3}x^{-2/3}}{-(3/4)a^{1/4}x^{-1/4}}$

as $\displaystyle x\rightarrow{a}$ we get:

=$\displaystyle \frac{0.5(2a^{3}a-a^4)^{-1/2}(2a^3-4a^3)-(1/3)a^{5/3}a^{-2/3}}{-(3/4)a^{1/4}a^{-1/4}}$

so

=$\displaystyle \frac{-a^{3}a^{-2}-a/3}{-{3/4}a}$

=$\displaystyle \frac{-a-a/3}{-(3/4)a}$

=$\displaystyle \frac{-(4/3)a}{-(3/4)a}$

=(4/3)/(3/4)

=16/9