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Evaluate limx→π62sin2x+sinx-12sin2x-3sinx+1

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प्रश्न

Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`

योग
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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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अध्याय 13: Limits and Derivatives - Solved Examples [पृष्ठ २३२]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 13 Limits and Derivatives
Solved Examples | Q 14 | पृष्ठ २३२

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