Continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$ If $f\colon\left[ { - 1,1} \right] \to \mathbb R$ be continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$, then $\mathop {\lim }\limits_{x \to 0} \frac{{f\left( {\cos x} \right)}}{{\sin x}}$ is _______.
My solution is as follow
$x = \cos \left( {\frac{\theta }{2}} \right)$
$f\left( {2{{\cos }^2}\left( {\frac{\theta }{2}} \right) - 1} \right) = \left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)\cos \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
How do I proceed from here
 A: First we note that $f$ is odd. The left hand side $f(2x^2-1)$ is even, and the factor $x^3+x$ in the right hand side is odd, so the other factor in the right hand side, $f(x),$ must also be odd. This implies that $f(0)=0.$
Taking $x=\sin\theta$ gives
$$
f(2\sin^2\theta - 1) = \sin\theta(\sin^2\theta+1) f(\sin\theta)
$$
so
$$
\frac{f(\cos 2\theta)}{\sin 2\theta}
= \frac{f(-(2\sin^2\theta - 1))}{2 \sin\theta \cos\theta}
= \frac{-\sin\theta(\sin^2\theta+1) f(\sin\theta)}{2 \sin\theta \cos\theta} \\
= -\frac{(\sin^2\theta+1) f(\sin\theta)}{2 \cos\theta}
\to -\frac{(0^2+1) f(0)}{2\cdot 1}
= 0
$$
as $\theta\to 0.$
A: Notice that $f(1)=0$. Since the function is continuous, $\lim_{x\to1}f(x)=0$

Also, $4xf'(2x^2-1)=(x^3+x)f'(x)+f(x)(3x^2+1)\implies f'(1)=\lim_{\theta\to0}f'(\cos(\theta/2))=0$

Now
$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)=\frac00$$
since $\lim_{\theta\to0}f(\cos(\theta/2))=\lim_{x\to1}f(x)=0$
Therefore by L'Hopital's
$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left(\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)\right)'}}{{\left(2\sin \left( {\frac{\theta }{2}} \right)\right)'}}=\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f'\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)}\frac{\sin(\theta/2)}{2}+f(\cos(\theta/2))\cos(\theta/2)\sin(\theta/2)}{{\cos \left( {\frac{\theta }{2}} \right)}}=??$$
