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
lab bhattacharjee
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Is this $\lim_{x\to 0}\dfrac{x^3}{x-\sin x}$=$\lim_{x\to 0}\dfrac{x-\sin x}{x^3}$ i true? – 1ENİGMA1 Oct 17 '18 at 07:23
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@1ENİGMA1, What do you think? – lab bhattacharjee Oct 17 '18 at 07:25
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@1ENİGMA1, But one thing is clear: if you know one,, you know the other? – lab bhattacharjee Oct 17 '18 at 07:30
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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 $\frac{x-\sin x}{x^3}$.
user
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