Need some help with L'Hopital's.

**qleeq** · March 10th 2011, 07:33 AM

$lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}$

Really no idea where to start with this. I think what's throwing me off is the a's. I know that they are constants but having a bit of trouble figuring out how to deal with them.

I can get this far.. and I'm not even sure if it's right or not.

$lim_{x \to a} \dfrac{(2a^3x-x^4)^{1/2} - a(a^2x)^{1/3}}{a - (ax^3)^{1/4}}$

$lim_{x \to a} \dfrac{\dfrac{2a^3-4x^3}{\sqrt{2a^3x-x^4}}-\dfrac{a^3}{3(a^2x)^{3/4}}}{-\dfrac{3\sqrt[4]{ax^3}}{4x}}$

**Sambit** · March 10th 2011, 07:56 AM

Your problem title says about L'Hospital's rule which needs differentiation, and then you need to know how to differentiate. If you are stuck with the constants (ie. the [i]a[/a]'s), note that $f'(ax)=af'(x)$

**qleeq** · March 10th 2011, 08:10 AM

Do I even need L'Hopital's method for this? I know it was the equation published in his book but still..

**Sambit** · March 10th 2011, 10:14 AM

Recheck the differentiation for last term in numerator. The power should be 2/3; not 3/4. You can then put a in place of x.

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