Find the limit $$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}$$
I found the limit which is $-\frac{1}{6}$ by using L'Hopital Rule. Is there another way to solve it without using the rule? Thanks in advance.
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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