I have a question about limit of Bessel function.
$$ \lim_{x \to \infty} x \bigg [ (J_p (x) )^2 + (Y_p (x) )^2 \bigg ] $$
Where, $ J_p (x)$ is Bessel function of first kind $ Y_p (x)$ is general solution of Bessel function.
I start like this from the definition and results of Bessel function of the first kind. Below first two are standard results for Bessel function for large $x$.
$$ \lim_{x \to \infty} J_p (x) \approx \sqrt{\frac{2}{x\pi }} cos( x - \frac{pi}{4} - \frac {p\pi}{2}) $$
$$ \lim_{x \to \infty} Y_p (x) \approx \sqrt{\frac{2}{x\pi }} sin( x - \frac{pi}{4} - \frac {p\pi}{2}) $$
$$ \implies \lim_{x \to \infty} \bigg [ x \frac{2}{x\pi}cos^2( x - \frac{pi}{4} - \frac {p\pi}{2}) + x \frac{2}{x\pi}sin^2( x - \frac{pi}{4} - \frac {p\pi}{2}) \bigg ] $$
$$ \implies \lim_{x \to \infty} \bigg [ \frac{2}{\pi}cos^2( x - \frac{\pi}{4} - \frac {p\pi}{2}) + \frac{2}{\pi}sin^2( x - \frac{\pi}{4} - \frac {p\pi}{2}) \bigg ] $$
Now, $sin(x),cos(x)$ are bounded between $[-1,1]$ So, what this limit becomes?
edit $$ \frac{2}{\pi} $$
Seems correct?