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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

Jochen
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Samar Imam Zaidi
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  • $f$ has a root at $-1$, from which follows that there is a root at $1$, for what it’s worth – FShrike Aug 17 '21 at 11:48
  • I found that by inspecting: $$f(\cos(2\theta))=(\cos^3(\theta)+\cos(\theta))f(\cos(\theta))$$ - Set $\theta=\pi/2$. – FShrike Aug 17 '21 at 11:51
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    And $\lim_{x\to0}f(\cos(x))=f(\lim_{x\to0}\cos(x))=f(1)=0$, which may also be helpful – FShrike Aug 17 '21 at 11:58
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    I think we can conclude that $f$ is also an odd function – FShrike Aug 17 '21 at 12:41
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    Let $p( \theta) = \frac{ f( \cos \theta) }{ \sin \theta}$, you have found:

    $$ p( \theta) = p( \frac{\theta}{2}) \frac{ \cos^2 \theta +1}{2}$$

    By induction, we find:

    $$ p(\theta) = \lim_{n \to \infty} p( \frac{\theta}{2^n})( \frac{\cos^2 \theta+1}{2})( \frac{\cos^2 \frac{\theta}{2} +1 }{2} )............ = \left[\lim_{t \to0} p(t) \right] \lim_{n \to \infty}\prod_{i=0}^{n}( \frac{\cos^2 \frac{ \theta}{2^i} +1}{2})$$

     – tryst with freedom Aug 17 '21 at 13:06

2 Answers2

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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 $\theta\to 0.$

md2perpe
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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

$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left(\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)\right)'}}{{\left(2\sin \left( {\frac{\theta }{2}} \right)\right)'}}=\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f'\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)}\frac{\sin(\theta/2)}{2}+f(\cos(\theta/2))\cos(\theta/2)\sin(\theta/2)}{{\cos \left( {\frac{\theta }{2}} \right)}}=??$$

DatBoi
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