`lim_(x rarr0)(sin^(2)x)/(x)= `

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`lim_(x rarr0)(sin^(2)x)/(x)= `

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Maheshbaliwal · Answer 1 · 2022-08-21T06:39:49+0000

This is 0/0 form

Apply L Hospital rule

lim(x→0) {d/dx(sin^2 x)}/(dx/dx)

lim(x→0) 2sinxcosx/1

lim(x→0) sin2x

Apply limit

Sin0=0

Option A is correct

`lim_(x rarr0)(sin^(2)x)/(x)= `

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