A very easy maths , asymptotic behavior

[VHAHAHA]

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 want to work it out
thank you

 

[voko]

Yes, the question is about the behavior of the function as x grows infinitely.

If you are unsure about your steps, show them here.

 

[VHAHAHA]

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?

 

[ehild]

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?

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

 

[VHAHAHA]

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 ?

 

[ehild]

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 ?

Yes, it is correct.

ehild