I want to find the limits limx→π/2cosxx−π/2
and
limx→π/4cotx−1x−π/4
and
limh→0sin2(π/4+h)−12h
without L'Hospital's Rule.
I know the fundamental limits limx→0sinxx=1,limx→0cosx−1x=0
Progress
Using cosx=sin(π2−x) I got −1 for the first limit.
I want to find the limits limx→π/2cosxx−π/2
and
limx→π/4cotx−1x−π/4
and
limh→0sin2(π/4+h)−12h
without L'Hospital's Rule.
I know the fundamental limits limx→0sinxx=1,limx→0cosx−1x=0
Using cosx=sin(π2−x) I got −1 for the first limit.
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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