Find limit as x-> a of this function

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  • #1
mathwiz123
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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?
 

Answers and Replies

  • #2
Saketh
261
2
I just want to make sure - is this the problem?
[tex]
y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{ax^3}}
[/tex]
 
  • #3
mathwiz123
10
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l'hopital

The diviser should be [a-(ax^3)^.25]
 
  • #4
stunner5000pt
1,447
2
[tex]
y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{\sqrt{ax^3}}}
[/tex]
 
  • #5
mathwiz123
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and instead of a^3(a*a*x), it should be a(a*a*x)^(1/3)
 
  • #6
stunner5000pt
1,447
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i wonder if flat out differentiaon would just do it...

maybe you rationalize the numerator?
 
  • #7
mathwiz123
10
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I think a little simplification after rationalization should do...hmmm. By the way, how do you do that math font?
 
  • #9
courtrigrad
1,236
1
[tex]
y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2x}} \right )}{a - \sqrt[4]{ax^3}}
[/tex]
 
  • #10
mathwiz123
10
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Thanks! And just for clarification. This is what I was trying to find.
[tex]y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}}[\tex]
 
  • #11
courtrigrad
1,236
1
[tex] y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}} [/tex]
 
  • #12
mathwiz123
10
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Yes! exactly. Thanks for the help.
 
  • #13
dextercioby
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The denominator can be brought to the form

[tex] a^4 -ax^3 [/tex]

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

Daniel.
 

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