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Homework Help: Find limit as x-> a of this function

  1. Oct 2, 2006 #1
    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. jcsd
  3. Oct 3, 2006 #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]
     
  4. Oct 3, 2006 #3
    l'hopital

    The diviser should be [a-(ax^3)^.25]
     
  5. Oct 3, 2006 #4
    [tex]
    y = \frac{\left ( \sqrt{2a^3x-x^4}-a^3\sqrt{a^{2x}} \right )}{a - \sqrt{\sqrt{ax^3}}}
    [/tex]
     
  6. Oct 3, 2006 #5
    and instead of a^3(a*a*x), it should be a(a*a*x)^(1/3)
     
  7. Oct 3, 2006 #6
    i wonder if flat out differentiaon would just do it...

    maybe you rationalize the numerator?
     
  8. Oct 3, 2006 #7
    I think a little simplification after rationalization should do...hmmm. By the way, how do you do that math font?
     
  9. Oct 3, 2006 #8
  10. Oct 3, 2006 #9
    [tex]
    y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2x}} \right )}{a - \sqrt[4]{ax^3}}
    [/tex]
     
  11. Oct 3, 2006 #10
    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]
     
  12. Oct 3, 2006 #11
    [tex] y = \frac{\left ( \sqrt{2a^3x-x^4}-a\sqrt[3]{a^{2}x} \right )}{a - \sqrt[4]{ax^3}} [/tex]
     
  13. Oct 3, 2006 #12
    Yes! exactly. Thanks for the help.
     
  14. Oct 4, 2006 #13

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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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