Find the Limit of S(x)/4x^3 as x Approaches 0 using Fresnel Function

  • Thread starter Thread starter turbokaz
  • Start date Start date
  • Tags Tags
    Calculus
turbokaz
Messages
19
Reaction score
0

Homework Statement


The Fresnel function is given as S(x) = ∫sin(3πt^2)dt from 0 to x. Find the limit as x approaches 0 of S(x)/4x^3


Homework Equations





The Attempt at a Solution


I took the derivative of the S(x) function to be able to plug x in. I then used L'Hospital's rule after getting 0/0. I took the derivative a second time after getting 0/0 again. My final answer was π/2, which is wrong. The other answer choices are π/4, 3π/2, 1/2, and 1/4.

sin(3πx^2)/4x^3 ; took derivative of top and bottom and got (6πx)(cos(3πx^2)/12x. Plugged in 0 and got 0/0. Took derivative of top and bottom again and got, (cos3πx^2)(6π)+(6πx)(-sin(3πx^2)(6πx). Plugged in 0 and got π/2 as final answer. Where did I go wrong and what is the right answer?
 
Physics news on Phys.org
When you took the derivative of S(x) = ∫sin(3πt^2)dt and plugged x into get sin(3πx^2), that counts as differentiating right? Did you do the same to the denominator? You claimed you started with sin(3πx^2)/4x^3, but shouldn't you start with sin(3πx^2)/12x^2 ?
 
OMFG. Derivative of 4x^3 is NOT 12x. WOW...I HATE WHEN I MAKE STUPID MISTAKES...Answer is pi/4. FML. Lost five points on my homework because of THAT stupid carelessness.
 
:D. It's moments like these that make you less prone to error in the future :P, atleast I find.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top