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Question

If f(x)=sin2xtan2x(ex1)x2 ,x0k ,x=0 is continuous at x=0, then the absolute value of k is

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Solution

For f(x) to be continuous at x=0
limx0f(x)=f(0)
Hence, k=limx0sin2xtan2x(ex1)x2
k=limx0sin2xsin2xcos2x(ex1)x2k=limx0sin2x(cos2x1)cos2x(ex1)x2k=limx0[(sin2x2x)(2sin2xx2)(1cos2x)(xex1)2]k=1211(2)k=4|k|=4

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