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A very easy maths , asymptotic behavior

  1. Jan 25, 2013 #1
    i know this question should be very simple, but i just dun know how to do
    what is asymptotic behavior means? x-> infinite ?
    and does this question need to use binomial expansion? or mayb binomial approximation?
    i got the ans is 0 but i think my step are not correct
    any tips ? i wanna work it out
    thank you
     

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  2. jcsd
  3. Jan 25, 2013 #2
    Yes, the question is about the behavior of the function as x grows infinitely.

    If you are unsure about your steps, show them here.
     
  4. Jan 25, 2013 #3
    ops , i notice that i hv made a error in my steps =(
    and now i couldn't find the ans , but i believe it should be 0

    here is my step

    1/√ (x^2 +a^2)^5 - 1/√(x^2 - a^5)^5
    =[√(x^2 - a^5)^5 - √ (x^2 +a^2)^5 ] /√ (x^2 +a^2)^5 (x^2 - a^5)^5

    use (a-b)(a+b) = a^2 - b^2

    [(x^2 - a^5)^5 - (x^2 +a^2)^5] / [√ (x^2 +a^2)^5 (x^2 - a^5)^5] [√(x^2 - a^5)^5 + √ (x^2 +a^2)^5 ]

    and i stop here

    what should i do in next step? expa [(x^2 - a^5)^5 - (x^2 +a^2)^5] ?
    any tips?
     
  5. Jan 25, 2013 #4

    ehild

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    You do not need any trick for this problem as the function is the difference of two vanishing terms.

    F(x)=u(x)-v(x) What are the limits of both u and v? You know that the limit of difference is equal to the difference of the limits, if they exist.

    ehild
     
  6. Jan 25, 2013 #5
    the limit of 1/√ (x^2 +a^2)^5 = 0? 1/√ (x^2 -a^2)^5 =0 ?
    but how to proof 1/√ (x^2 +a^2)^5 = 0?
    because x tends infi, so (x^2 + a^2) ^5 -> infi and 1/ (x^2 +a^2)^5 --> 0 ?
     
  7. Jan 25, 2013 #6

    ehild

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    Yes, it is correct.

    ehild
     
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