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Can someone help me with a approach?
Not looking for a solution just for a hint
$\endgroup$ 53 Answers
$\begingroup$The only thing you need to do is prove that $\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=0$. Bring to the common denominator and use L'Hopital's rule (twice).
$\endgroup$ 6 $\begingroup$You could use Taylor series:${ 1\over \sin x } - { 1\over x} = {x - \sin x \over x \sin x} = { x- (x-{x^3 \over 3!} + \cdots ) \over x^2-{x^4 \over 3!} + \cdots } = {x^3 \over x^2} {{1 \over 3!} +\cdots\over 1 -{x^2 \over 4!}+\dots}$.
$\endgroup$ 2 $\begingroup$Hint
Both $\sin x$ and $x$ are continuous and non-zero over $(0,\pi)$. For checking $x=0$ simply use $$\sin x=x-{x^3\over 6}\quad,\quad |x|\ll1$$
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