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 🙂
Without using L’Hospital rule or series expansion find $\lim_{x\to0} \frac{x-x\cos x}{x-\sin x}$.
calculuslimitslimits-without-lhopital
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Best Answer
$$\dfrac{x(1-\cos x)}{x-\sin x}=\dfrac{x^3}{x-\sin x}\cdot\dfrac1{1+\cos x}\left(\dfrac{\sin x}x\right)^2$$
For $\lim_{x\to0}\dfrac{x^3}{x-\sin x}$ use Are all limits solvable without L'Hôpital Rule or Series Expansion