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Evaluate: limx→π63sinx-cosxx-π6 - Mathematics

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प्रश्न

Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`

योग
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उत्तर

Given that `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`

= `lim_(x -> pi/6) (2[sqrt(3)/2  sin x - 1/2 cos x])/(x - pi/6)`

= `lim_(x -> pi/6) (2[cos  pi/6  sin x - sin  pi/6  cos x])/(x - pi/6)`

= `lim_((x -> pi/6),(because  x - pi/6 -> 0)) (2sin (x - pi/6))/((x - pi/6))`  .......`[because  lim_(x -> 0) sinx.x = 1]`

= `2 * 1`

= 2

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अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 22 | पृष्ठ २४०

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