Monday, 6 October 2014

limits - limntoinfty2ncosleft(fracpi2nright)sinleft(fracpi2nright) (Without L'Hospital)

I'm trying to find
limn2ncos(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?

No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...