how can i compute this limit without L'hopital rule:
$\lim \limits _{z\to 0} \frac{1-cos z}{z^2}$
i believe the answer is 1/2 by the rule,
thanks
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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