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Question

Solve: limx0sin2(π2ax)sec2π2bx

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Solution

We have, limx0sin2(π2ax)sec2π2bx

=elimx0cos2(π/(2ax))cos2(π/2bx)

=elimx0sin2(π/2π/(2ax))sin2(π/2π/(2bx))

=elimx0sin2(πax/2(2ax))sin2(πbx/2(2bx))=e4a24b2=ea2/b2


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