Is it possible to find limx→0x−xcosxx−sinx without using L'Hopital's Rule or Series expansion.
I can't find it.If it is dublicated, sorry :)
Answer
x(1−cosx)x−sinx=x3x−sinx⋅11+cosx(sinxx)2
For limx→0x3x−sinx use Are all limits solvable without L'Hôpital Rule or Series Expansion
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