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Question

limx01x(cyesin2tdtcx+yesin2tdt) is equal to (Where c is a constant)

A
esin2y
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B
sin2yesin2t
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C
0
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D
None of these
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Solution

The correct option is A esin2y
limx01x(cyesin2tdtcx+yesin2tdt)=limx0x+yyesin2tdtx

According to L'hospital's Rule,
limx0esin2(x+y)(1+dydx)esin2ydydx
limx0esin2(x+y)+(limx0esin2(x+y)esin2yx)limx0dydxx
esin2y+0

So, Option A is the correct answer.

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