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

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

Grade:11

2 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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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Badiuddin

Bhanu Singh Tanwar
13 Points
2 years ago
limx->0 (sin2x-x2)/x4 =
limx->0 (2sinxcosx-2x)/4x3
limx->0 (sinxcosx-x)/2x3
limx->0 (cox2x-sin2x-1)/6x2   ; (cos2x-1= -sin2x)
limx->0 (-2sin2x)/6x2= -1/3
 

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