$$\lim_{x\to\ 0}\;x \tan\left(\dfrac{{\pi}\left(1-x\right)}{2}\right)$$
How do you do this without L'Hospital's Rule?
The result is supposed to be $\dfrac{2}{{\pi}}$.
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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