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