Question

lim x-0 1/x² - 1/sin²x
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Answer 1

Answer:

-1/3

Step-by-step explanation:

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

Answer 2

Answer is raeght

sahi hi