मराठी

Evaluate: limx→π31-cos6x2(π3-x)

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प्रश्न

Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`

बेरीज
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उत्तर

Given that `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`

= `lim_(x -> pi/2) (sqrt(2 sin^2 3x))/(sqrt(2) (pi/3 - x))`  ......`[because 1 - cos theta = 2 sin^2  theta/2]`

= `lim_(x -> pi/3) (sqrt(2) sin 3x)/(sqrt(2)((pi - 3x)/3))`

= `lim_((x -> pi/3),(because  pi - 3x -> 0)) (3 * sin (pi - 3x))/(pi - 3x)`

= 3   .....`[because  lim_(x -> 0)  sinx/x = 1]`

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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 20 | पृष्ठ २४०

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