Find limit as x-> a of this function

[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?

 

[Saketh]

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

 

[mathwiz123]

l'hopital

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

 

[stunner5000pt]

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

 

[mathwiz123]

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

 

[stunner5000pt]

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

maybe you rationalize the numerator?

 

[mathwiz123]

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

 

[stunner5000pt]

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

https://www.physicsforums.com/misc/howtolatex.pdf

 

[courtrigrad]

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

 

[mathwiz123]

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

 

[courtrigrad]

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

 

[mathwiz123]

Yes! exactly. Thanks for the help.

 

[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.