Q.
The value of the limit limθ→0tanπcos2θsin2πsin2θ is equal to:
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a
−12
b
14
c
−14
d
0
answer is A.
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Detailed Solution
limθ→0tanπcos2θsin2πsin2θ=limθ→0tanπcos2θsin2π−2πcos2θ=limθ→0tanπcos2θsin2π−2πcos2θ=−limθ→0tanπcos2θsin2πcos2θ=−limθ→0tanπcos2θ2sinπcos2θcosπcos2θ=−12limθ→0sec2πcos2θ=−12sec20=−12
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