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lim (sin2x)tan^2(2x)
x_pie/4

ashu , 11 Years ago
Grade 11
anser 1 Answers
Sher Mohammad
1infiniteform
lim 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 form
e^(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 Mohammad
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Last Activity: 11 Years ago
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