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Question: Answered & Verified by Expert
The value of limxπ4sin2xsec22x is equal to
MathematicsLimitsJEE Main
Options:
  • A -12
  • B 12
  • C e-12
  • D e12
Solution:
2297 Upvotes Verified Answer
The correct answer is: e-12
 limxπ4sin2xsec22x 

=elimxπ4sec22xsin2x-1=elimxπ4sin2x-1cos22x

Applying L'Hospital Rule, we get, 

=elimxπ42cos2x2cos2x-2sin2x

=elimxπ4-12sin2x=e-12

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