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प्रश्न
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
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उत्तर
Note that `2sin^2x + sin x - 1 = (2 sin x - 1)(sin x + 1)`
`2sin^2x - 3 sin x + 1 = (2 sin x - 1) (sin x - 1)`
Therefore, `lim_(x -> pi/6) (2sin^2x + sinx - 1)/(2sin^2x - 3sin x + 1)`
= `lim_(x -> pi/6) ((2sinx - 1)(sinx + 1))/((2sinx - 1)(sin x - 1))`
= `lim_(x -> pi/6) (sinx + 1)/(sinx - 1)` ......(As 2 sin x – 1 ≠ 0)
= `(1 + sin pi/6)/(sin pi/6 - 1)`
= – 3
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