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a
e2
b
e
c
1e
d
-1
answer is B.
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Detailed Solution
limx→0 1+2x3(tanx−sinx)−12/x2=elimx→02(tanx−sinx)−x3⋅2x5 =elimx→022x−x33+2x515−⋯−x−x33!+x55!⋯−x3x5=elimx→0 22x58+x7 and higher powers of xx5=e1/2=e