Is it possible to find lim without using L'Hopital's Rule or Series expansion.
I can't find it.If it is dublicated, sorry :)
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
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