Given the fact: (Asymptotic form) For $\theta \in (0, \pi)$, we have $$ P_n(\cos \theta)= J_0(n\theta)+ O(\dfrac{1}{n})= \sqrt{\dfrac{2}{\pi n \sin \theta}}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n}),$$ (where $J_0$ is the zeroth Bessel function of the first kind). Consequently, for $x \in [-1,1]$ we get $\lim_{n \rightarrow \infty} P_n(x) = 0.$
Prove that for $x \in [-1,1]$ and $n \geq 1$, $$ (1-x^2)^{\frac{1}{4}} \lvert {P_n(x)} \rvert < \sqrt{\dfrac{2}{n \pi}}.$$
My approach:
Let $$x=\cos\theta$$. $$ \implies x^2 = \cos^2 \theta$$ $$ \implies x^2 = 1 - \sin^2 \theta$$ $$ \implies sin^2 \theta = x^2 - 1$$ $$ \implies \sqrt{sin\theta}= (x^2-1)^{\frac{1}{4}}$$ \ Given, $$ P_n(\cos \theta)= \sqrt{\dfrac{2}{\pi n \sin \theta}}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n}),$$ $$ \sqrt{sin\theta}P_n(\cos \theta) = \sqrt{sin\theta}\sqrt{\dfrac{2}{\pi n \sin \theta}}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n})$$ $$ \sqrt{sin\theta}P_n(\cos \theta) = \sqrt{\dfrac{2}{\pi n }}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n})$$ $$ \sqrt{sin\theta}P_n(\cos \theta) = \sqrt{\dfrac{2}{\pi n }}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n})$$ $$ (x^2-1)^{\frac{1}{4}}P_n(\cos \theta) = \sqrt{\dfrac{2}{\pi n }}\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right) + O(\dfrac{1}{n})$$
Since $\sin \left(\dfrac{2n+1}{2}\theta + \dfrac{\pi}{4}\right)$ is bounded between -1 and 1 with a max value of 1, and $ O(\dfrac{1}{n})$ is an error which is negligible, how would $ (1-x^2)^{\frac{1}{4}} \lvert {P_n(x)} \rvert < \sqrt{\dfrac{2}{n \pi}}.$? Isn't it $ (1-x^2)^{\frac{1}{4}} \lvert {P_n(x)} \rvert \leq \sqrt{\dfrac{2}{n \pi}}.$ instead?