Find limit as x-> a of this function

1. Oct 2, 2006

mathwiz123

This is one of the actual example L'hopital used in his book "L'analyse des Infiniment Petits Pour I'Intelligence des Lignes Courbes"

Check it out!

Find the limit of x as it approaches a for

y=[[(2*a^3*x-x^4)^.5]-[a^3(a^2x)^.5]]/(a-(ax^3)^.25)

Unfortunately, I can't get rid of the indeterminate form! Ideas/help?

2. Oct 3, 2006

Saketh

I just want to make sure - is this the problem?
$$y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{ax^3}}$$

3. Oct 3, 2006

mathwiz123

l'hopital

The diviser should be [a-(ax^3)^.25]

4. Oct 3, 2006

stunner5000pt

$$y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{\sqrt{ax^3}}}$$

5. Oct 3, 2006

mathwiz123

and instead of a^3(a*a*x), it should be a(a*a*x)^(1/3)

6. Oct 3, 2006

stunner5000pt

i wonder if flat out differentiaon would just do it...

maybe you rationalize the numerator?

7. Oct 3, 2006

mathwiz123

I think a little simplification after rationalization should do...hmmm. By the way, how do you do that math font?

8. Oct 3, 2006

stunner5000pt

9. Oct 3, 2006

$$y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2x}} \right )}{a - \sqrt[4]{ax^3}}$$

10. Oct 3, 2006

mathwiz123

Thanks! And just for clarification. This is what I was trying to find.
$$y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}}[\tex] 11. Oct 3, 2006 courtrigrad [tex] y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}}$$

12. Oct 3, 2006

mathwiz123

Yes! exactly. Thanks for the help.

13. Oct 4, 2006

dextercioby

The denominator can be brought to the form

$$a^4 -ax^3$$

which is easy to differentiate when using the 'H^opital rule.

Daniel.