How to solve $\lim\limits_{x\to 3}\frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2\sin \frac{3π}{2}\sin\frac{πx}{2}\right)$ Without L'Hopital.

Question

I want to compute the following limit

$$\lim_{x\to 3}\frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2\sin \frac{3π}{2}\sin\frac{πx}{2}\right)$$

But I am looking for a more elegant way not using L'Hopital rule. especially squeezing will be good and welcome.

Answer 1

Let $x = y+3$, so $x^2=y^2+6y+9$. Then

$\begin{array}\\ \frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2\sin \frac{3π}{2}\sin\frac{πx}{2}\right) &=\frac{1}{(9-(y^2+6y+9))^2}\left( \frac{9+y^2+6y+9}{3(y+3)}+2\sin\frac{π(y+3)}{2}\right)\\ &=\frac{1}{(-y^2-6y)^2}\left( \frac{y^2+6y+18}{3y+9}+2\sin\frac{πy+3\pi}{2}\right)\\ &=\frac{1}{y^2(y+6)^2}\left( \frac{y^2+6y+18}{3y+9}+2(\sin(\pi y/2)\cos(3\pi/2)+\cos(\pi y/2)\sin(3\pi/2))\right)\\ &=\frac{1}{y^2(y+6)^2}\left( \frac{y^2+6y+18}{3y+9}+2(-\cos(\pi y/2))\right)\\ &=\frac{1}{y^2(y+6)^2}\left( \frac{y^2+6y+18}{3y+9}-2(1-(\pi y/2)^2/2+O(y^4))\right)\\ &=\frac{1}{y^2(y+6)^2}\left( \frac{y^2+6y+18-2(3y+9)}{3y+9}+\pi^2 y^2/+O(y^4)\right)\\ &=\frac{1}{y^2(y+6)^2}\left( \frac{y^2}{3y+9}+\pi^2 y^2/4+O(y^4)\right)\\ &=\frac{1}{(y+6)^2}\left( \frac{1}{3y+9}+\frac{\pi^2}{4}+O(y^2)\right)\\ &\to\frac{1}{36}\left( \frac{1}{9}+\frac{\pi^2}{4}\right)\\ \end{array} $

Answer 2

Hint:

Using Werner Formulas, $2\sin A\sin B=\cos(A-B)-\cos(A+B),$

$$\frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2\sin \frac{3π}{2}\sin\frac{πx}{2}\right)$$

$$\frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2+2+\cos\dfrac{\pi(3+x)}2-\cos\dfrac{\pi(3-x)}2\right)$$

$$=\dfrac1{(3+x)^2}\left(\dfrac{(3-x)^2}{3x(3-x)^2}+\dfrac{1+\cos\dfrac{\pi(3+x)}2}{(3-x)^2}+\dfrac{1-\cos\dfrac{\pi(3-x)}2}{(3-x)^2}\right)$$

Answer 3

Let $x=y+3$ with $y\to0$

$$\lim_{x\to 3}\frac{1}{(9-x^2)^2}\left( \frac{9+x^2}{3x}-2\sin \frac{3π}{2}\sin\frac{πx}{2}\right)=\lim_{y\to 0}\frac{1}{(-y^2-6y)^2}\left( \frac{y^2+6y+18}{3y+9}-2\cos \left(\frac{\pi}{2}y\right)\right)$$

then

$$\frac{1}{(y^2+6y)^2}\left( \frac{y^2+6y+18}{3y+9}-2+\frac{\pi^2}{4}y^2+o(y^2)\right)=\frac{1}{(y^2+6y)^2}\left( \frac{y^2+6y+18-6y-18}{3y+9}+\frac{\pi^2}{4}y^2+o(y^2) \right)=\frac{1}{(y+6)^2}\left( \frac{1}{3y+9}+\frac{\pi^2}{4}+o(1) \right)\to\frac1{36}\cdot\left(\frac19+\frac{\pi^2}{4}\right)$$