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प्रश्न
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
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उत्तर
`lim_(x -> pi/6) (2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)`
= `lim_(x -> pi/6) ((2sinx - 1)(sin x + 1))/((2sin x - 1)(sin x - 1))`
= `lim_(x -> pi/6) (sinx + 1)/(sin x - 1) ...[(because x -> pi/6"," x ≠ pi/6),(therefore sin x ≠ sin pi/6 = 1/2),(therefore 2 sin x - 1 ≠ 0)]`
= `(lim_(x -> pi/6) (sin x + 1))/(lim_(x -> pi/6) (sin x - 1))`
= `(sin pi/6 + 1)/(sin pi/6 - 1)`
= `(1/2 + 1)/(1/2 - 1)`
= `(1 + 2)/(1 - 2)`
= – 3
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