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				   lim x-0 (1\x square -1\sin square x)


6 years ago

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										Dear arun
lim x-0 (1/x2 -1/sin 2x)
= lim x-0 (sin 2x - x2)/x2sin 2x
=lim x-0 (sin 2x - x2)/x4(sin x/x)2
we know lim x-0  sin x/x   =1
so limit become

=lim x-0 (sin 2x - x2)/x4
=lim x-0 (1/2 -cos2x/2 - x2)/x4
=lim x-0 (1 -cos2x - 2x2)/2x4
=lim x-0 (1 - 2x2-cos2x)/2x4
open series of cos2x
=lim x-0 (1 - 2x2-[1-(2x)2/2!  + (2x)4/4! - (2x)6/6! +.......])/2x4
=lim x-0 (- (2x)4/4! +(2x)6/6! +.......])/2x4
lim x-0 (- 1/3  +32(x)2/6! -.......])
apply limit
=-1/3
Please 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. Regards,Askiitians ExpertsBadiuddin

6 years ago

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