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Question

Let [x] denote the greatest integer less than or equal to x. Then :
limx0tan(π sin2 x)+(|x|sin(x[x]))2x2:

A
equals π
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B
equals 0
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C
equals π+1
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D
does not exist
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Solution

The correct option is D does not exist
LHL=limx0tan(π sin2 x)+(xsin x(1))2x2

=limx0tan(πsin2 x)π sin2 x×π sin2 xx2+(1+sinxx)2

=1×π+(1+1)2
=π


RHL=limx0+tan(π sin2 x)+(xsinx2)2x2
=limx0+tan(π sin2 x)πsin2 x×πsin2 xx2+(1sinx2x)2

=1×π+(10)2
=π+1

LHL RHL
Limit does not exist.

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