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Is it possible to find $\displaystyle{\lim_{x\to 0} \frac{x-x\cos x}{x-\sin x}}$ without using L'Hopital's Rule or Series expansion.


I can't find it.If it is dublicated, sorry :)

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$$\dfrac{x(1-\cos x)}{x-\sin x}=\dfrac{x^3}{x-\sin x}\cdot\dfrac1{1+\cos x}\left(\dfrac{\sin x}x\right)^2$$

For $\lim_{x\to0}\dfrac{x^3}{x-\sin x}$ use Are all limits solvable without L'Hôpital Rule or Series Expansion

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  • $\begingroup$ Is this $\lim_{x\to 0}\dfrac{x^3}{x-\sin x}$=$\lim_{x\to 0}\dfrac{x-\sin x}{x^3}$ i true? $\endgroup$ – 1ENİGMA1 Oct 17 '18 at 7:23
  • $\begingroup$ @1ENİGMA1, What do you think? $\endgroup$ – lab bhattacharjee Oct 17 '18 at 7:25
  • $\begingroup$ @1ENİGMA1, But one thing is clear: if you know one,, you know the other? $\endgroup$ – lab bhattacharjee Oct 17 '18 at 7:30
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We have that

$$\frac{x-x\cos x}{x-\sin x}=\frac{\frac{x-x\cos x}{x^3}}{\frac{x-\sin x}{x^3}}=\frac{\frac{1-\cos x}{x^2}}{\frac{x-\sin x}{x^3}}$$

then refer to standard limit $\frac{1-\cos x}{x^2}\to \frac12$ and to the link already given for $\frac{x-\sin x}{x^3}$.

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