मराठी

Evaluate: limx→02-1+cosxsin2x - Mathematics

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प्रश्न

Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`

बेरीज
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उत्तर

Given that `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`

= `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x) xx (sqrt(2) + sqrt(1 + cosx))/(sqrt(2) + sqrt(1 + cos x))`

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

= `lim_(x -> 0) (1 - cos x)/(sin^2 xx [sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) (2 sin^2  x/2)/((2 sin  x/2 cos  x/2)^2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) (2 sin^2  x/2)/(4 sin^2  x/2 cos^2  x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

= `lim_(x -> 0) 2/(4 cos^2  x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`

Taking limit, we get

= `2/(4 cos^2 0) xx 1/((sqrt(2) + sqrt(2))`

= `1/2 xx 1/(2sqrt(2))`

= `1/(4sqrt(2))`

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

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 26 | पृष्ठ २४०

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