I can't figure out the next equation. The answer is negative infinity, but i don't know how to get there by using L'Hospital.The equation is:
$$\lim_{x\to 0} \frac{x-\sin(2x)}{x^3}$$
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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