$f:[-1,1]\to \mathbb{R}$ is a continuous function such that $f(2x^2-1)=(x^3+x)f(x).$ Find $\lim_{x\to 0}\frac {f(\cos x)}{\sin x}$

Question

$f:[-1,1]\to \mathbb{R}$ is a continuous function such that $f(2x^2-1)=(x^3+x)f(x).$ Find $\lim_{x\to 0}\frac {f(\cos x)}{\sin x}$

My approach : Using the functional equation I got f is an odd function. $\implies f(0)=0$

let $g(x)=\frac{f(\cos x)}{\sin x},x\ne 0\implies |g(2x)|= \frac{1+\cos^2x}{2}|g(x)|\le |g(x)|$ $\implies |g(x)|\le |g\left(\frac {x}{2^n}\right)|...(1)$

Also By replacing $x$ by $\sin x$ in the given functional equation we get $g(2x)=-\frac{1+\sin^2x}{2}\frac{f(\sin x)}{cos x}...(2)$ From (1) and (2) we get, $|g(x)|\le \left|\frac {1+\sin^2 \frac{x}{2^{n+1}}}{2\cos \frac{x}{2^{n+1}}}f\left(\sin\frac {x}{2^{n+1}}\right)\right|$

lets $n\to\infty\implies |g(x)|\le 0\implies g(x)=0\,\,\forall x\in[-1,1]-\{0\}.$

$\implies \lim_{x\to 0}\frac {f(\cos x)}{\sin x}=0$

Answer 1

Also, $$f(1)=2f(1),$$ which gives $$f(-1)=f(1)=0.$$ Now, for $x=\cos\frac{\pi}{4}$ we obtain $$f\left(\cos\frac{\pi}{4}\right)=0,...$$ which gives $$f\left(\cos\frac{k\pi}{2^n}\right)=0$$ for all $n\in\mathbb N$ and $k\in\mathbb Z$, $|k|\leq 2^{n}.$

Thus, since $f$ is a continuous function, we obtain $f(x)=0$ for all $x\in[-1,1],$

which says that our limit is equal to $0$.