Evaluate
limx→π/2√1+cos(2x)√π−√2x
I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
No comments:
Post a Comment