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Question

Evaluate limx0ex2cosxx2 is mn. Find m+n

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Solution

Ltx0ex2cosxx2=Lx0t(ex21)+(1cosx)x2
=Ltx0ex21x2+Lx0t2sin2x2x2
=loge+2.(x/2)2x2=1+12=32
m+n=3+2=5

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