`lim_(x to pi//6) (2sin^2x+sinx-1)/(2sin^2x-3sinx+1)=`

Correct Answer - B
We have,
` lim_(x to pi//6) (2sin^2x+sinx-1)/(2sin^2x-3sinx+1)`
`=lim_(x to pi//6)((2sinx-1)(sinx+1))/((2sinx-1)(sinx-1))`
`=lim_(x to pi//6) (sin x +1)/(sinx-1)=-3`

`lim_(x to pi//6) (2sin^2x+sinx-1)/(2sin^2x-3sinx+1)=`

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