Without using L'Hospital rule or series expansion find $\lim_{x\to0} \frac{x-x\cos x}{x-\sin x}$.

Question

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.

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

Answer 1

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

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

Answer 2

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 $\frac{x-\sin x}{x^3}$.