I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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