Find the limits without L'Hôpital's rule
limx→0x−sinxx−tanx=?
My Try:
limx→0sin(π−x)−sinxtan(π+x)−tanx=?limx→02sin(π2−x)cos(π2)sin(π2−x)cos(π+x)cos(x)=limx→0(2cosx)(−cosx)(cos(π2))cosx=0
but:
limx→0x−sinxx−tanx=−1/2
Where is my mistake?
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
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