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Limx→π1-sin x2cos x2(cos x4-sin x4) - Mathematics

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प्रश्न

`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`

योग
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उत्तर

Given, `lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`

= `lim_(x -> pi) (cos^2  x/4 + sin^2  x/4 - 2 sin  x/4 * cos  x/4)/((cos^2  x/4 - sin^2  x/4)(cos  x/4 - sin  x/4))`  ......`[because  cos 2theta = cos^2theta - sin^2theta]`

= `lim_(x -> pi) (cos  x/4 - sin  x/4)^2/((cos  x/4 - sin  x/4) * (cos  x/4 + sin  x/4) * (cos  x/4 - sin  x/4))`

= `lim_(x -> pi) 1/((cos  x/4 + sin x x / 4))`

Taking limits we have

= `1/(cos  pi/4 + sin  pi/4)`

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

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

= `1/sqrt(2)`

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अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 50 | पृष्ठ २४१

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