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lim square root[1 - square root(sin 2x)]
x -- pi/4 pi - 4x
Dear iit jee , square root[1 - square root(sin 2x)] pi/4 - xmultiplying both numerator and denominator with sq.rt (1+sq.rt sin2x)we getmod(cosx-sinx) /sq.rt(1+sq.rt sin 2x)so .., lim square root[1 - square root(sin 2x)] x -- pi/4 pi/4 - x=lim mod(cosx-sinx) /sq.rt(1+sq.rt sin 2x)*(pi-4x) x->pi/4mod(cosx-sinx)= cosx-sinx for x->(pi/4)- = sinx-cosx for x->(pi/4)+ lim square root[1 - square root(sin 2x)] x -- pi/4 pi/4 - x= lim sinx-cosx/ sq.rt(1+sq.rt sin 2x)*(pi-4x) for x->(pi/4)+= infinity lim tends to 1 when x->pi/4-hence the function is discontinous at pi/4Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best iit jee !!!Regards,Askiitians ExpertsChaitanya Sagar
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