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

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In summary, the conversation discusses finding the limit of the Fresnel function S(x) as x approaches 0, which is given by S(x) = ∫sin(3πt^2)dt from 0 to x, divided by 4x^3. The student attempted to solve it using L'Hospital's rule but made a mistake in taking the derivative of the denominator. The correct answer is π/4.
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turbokaz
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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?
 
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  • #2
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 ?
 
  • #3
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.
 
  • #4
:D. It's moments like these that make you less prone to error in the future :P, atleast I find.
 

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

What is the Fresnel function?

The Fresnel function, denoted by S(x), is a special function in mathematics that arises in the study of optics and wave phenomena. It is defined as the integral of sine and cosine functions from 0 to x.

Why is finding the limit of S(x)/4x^3 important?

Finding the limit of S(x)/4x^3 is important in calculus as it helps determine the behavior of a function as the variable (in this case, x) approaches a certain value (in this case, 0). It is also important in applications involving wave phenomena and optics.

How do you find the limit of S(x)/4x^3 as x approaches 0?

To find the limit of S(x)/4x^3 as x approaches 0, we can use L'Hôpital's rule which states that the limit of a quotient of two functions is equal to the limit of the derivatives of those functions, provided that the limit of the derivative of the denominator is not 0.

Can the limit of S(x)/4x^3 as x approaches 0 be evaluated without using L'Hôpital's rule?

Yes, it is possible to evaluate the limit of S(x)/4x^3 as x approaches 0 without using L'Hôpital's rule. One method is to use Taylor series expansion to approximate the function and then take the limit as x approaches 0.

What is the significance of the limit of S(x)/4x^3 as x approaches 0?

The limit of S(x)/4x^3 as x approaches 0 has physical significance in the study of wave phenomena and optics. It represents the behavior of a wave as it approaches a boundary or obstacle, and is used to calculate parameters such as reflection and refraction coefficients. It also has applications in diffraction and interference phenomena.

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