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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