Taks: $$\lim_{x \to 0} \frac{2 \sin x-\sin 2x}{x-\sin x}$$
Is it possible to solve it without l'Hopital ?
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Often you can use power series approximations. Here, use: $$\sin(x)=x-\frac{x^3}{6}+o(x^3)$$ – Thomas Andrews Jun 11 '21 at 15:38
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See this post – M. Strochyk Jun 11 '21 at 15:42
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4Does this answer your question? Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule – 311411 Jun 11 '21 at 15:50