I'm trying to find
limn→∞2ncos(π2n)sin(π2n)
I think the answer is π, but I don't know how to find it.
Could you please show me the shortcut?
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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