1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

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

    Attached Files:

  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


    User Avatar
    Homework Helper

    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.

  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


    User Avatar
    Homework Helper

    Yes, it is correct.

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