मराठी

Evaluate: limx→asinx-sinax-a

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प्रश्न

Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`

बेरीज
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उत्तर

Given that `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`

= `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a)) xx (sqrt(x) + sqrt(a))/(sqrt(x) + sqrt(a))`

= `lim_(x -> a) ((sin x - sin a)(sqrt(x) + sqrt(a)))/(x - a)`

= `lim_(x -> a) ((2 cos  (x + a)/2 * sin  (x - a)/2)(sqrt(x) + sqrt(a)))/(x - a)`

= `lim_((x -> a),(because  (x - a)/2 -> 0)) (2 cos  (x + a)/2 * (sin  (x - a)/2)/(2 xx (x - a)/2)) (sqrt(x) + sqrt(a))`

= `lim_(x -> a) cos((x + a)/2)(sqrt(x) + sqrt(1))`  .....`[because  lim_((x - a)/2 -> 0)  (sin  (x - a)/2)/((x - a)/2) = 1]`

Taking limit we have

= `cos ((a + a)/2)(sqrt(a) + sqrt(a))`

= `cos a xx 2sqrt(a)`

= `2sqrt(a) * cos a`

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

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 24 | पृष्ठ २४०

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