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

6 years ago


Answers : (1)


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


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6 years ago

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