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Question:

Let [x] denote the integer less than or equal to x. Then: limx→0[tan(πsin2x) + (|x| - sin(x[x]))] / 2x²

equals 0

does not exist

equals π

equals π + 1

Solution:

The correct option is D does not exist
R.H.L=limx→0+[tan(πsin2x) + (|x| - sin(x[x]))]/2x² (as x→0+ ⇒ [x]=0)
limx→0+[tan(πsin2x)]/πsin2x + 1 = π + 1
L.H.L.=limx→0-[tan(πsin2x) + (-x + sinx)]/2x² (as x→0- ⇒ [x]=-1)
limx→0+[tan(πsin2x)]/πsin2x. πsin2x/x² + (-1 + sinx)/x² → ∞
R.H.L ≠ L.H.L.