Question

The Fresnel function S(x) = integral 0 to x sin (9/2 pi t^2) dt is important in the theory of Fourier Optics in Physics. Find the value of (Hint: remember L'Hospital's Rule.)

1. limit = 3/5
2. limit = 9/5 pi
3. limit = 3/10
4. limit = 3/10 pi
5. limit = 3/5 pi

Answer 1

Answer:

$\frac{3\pi}{10}$

Step-by-step explanation:

$S(x) = \int\limits_0^x \sin (\frac{9}{2} \pi t^2) dt\\\\\lim\limits_{x \to 0} \frac{S(x)}{5x^3} =\lim\limits_{x \to 0} \frac{\int\limits_0^x \sin (\frac{9}{2} \pi t^2) dt}{5x^3}$

We will use L'Hospital Rule twice.

Firstly, to find the derivative of the integral, we will use Fundamental Theroem of Calculus.

$\frac{d}{dx} \lim\limits_{x \to 0} \int\limits_0^x \sin (\frac{9}{2} \pi t^2) dt= \sin (\frac{9}{2} \pi x^2)$

Hence,

$S(x) = \int\limits_0^x \sin (\frac{9}{2} \pi t^2) dt\\\\\lim\limits_{x \to 0} \frac{S(x)}{5x^3} =\lim\limits_{x \to 0} \frac{\int\limits_0^x \sin (\frac{9}{2} \pi t^2) dt}{5x^3}=\lim\limits_{x \to 0} \frac{\sin (\frac{9}{2} \pi x^2) }{15x^2} =\lim\limits_{x \to 0} \frac{9\pi x \cos(\frac{9}{2} \pi x^2)}{30x} =\\\\=\lim\limits_{x \to 0} \frac{9\pi \cos(\frac{9}{2} \pi x^2)}{30}=\frac{9\pi}{30} =\frac{3\pi}{10}$