How to find limh→0sin(ha)h without lhopital rule?
I know when I use lhopital I easy get
limh→0cos(ah)a1=a but I don't know how to behave without that way
Answer
Hint:
sin(ha)h=a⋅sin(ha)ha
Also, remember what limx→0sin(x)x is equal to?
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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