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lim (sin2x)tan^2(2x)x_pie/4
1infiniteformlim x_pie/4 (1+sin2x-1)^tan^2(2x)=e^(lim x_pie/4 (sin2x-1)*(tan^2(2x)))apply L Hospital rule because 0/0 forme^(lim x_pie/4 (sin2x-1)/(cot^2(2x))=e^(lim x_pie/4 (2*cos2x)/(4*cot2x*cosec^2(2x))=e^(lim x_pie/4 (2*cos2x*sin^3(2x))/(4*cos2x)=e^(1/2)Sher Mohammadfaculty askiitians
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