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

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 :)

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

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

• 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 7:23
• @1ENİGMA1, What do you think? – lab bhattacharjee Oct 17 '18 at 7:25
• @1ENİGMA1, But one thing is clear: if you know one,, you know the other? – lab bhattacharjee Oct 17 '18 at 7:30

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