1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?