limx→0ln(cosx)sin2x=?
I can solve this by using L'Hopital's rule but how would I do this without this?
Answer
log(cos(x))sin2(x)=12log(1−sin2(x))sin2(x)=−sin2(x)+O(sin4(x))2sin2(x)x→0→−12.
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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