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Evaluate the following limit : limx→π4[cosx-sinxcos2x]

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प्रश्न

Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`

योग
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उत्तर

`lim_(x -> pi/4) (cosx - sinx)/(cos2x)`

= `lim_(x -> pi/4) (cosx - sinx)/(cos^2x - sin^2x)`

= `lim_(x -> pi/4) (cosx - sinx)/((cosx - sinx)(cosx + sinx))`

= `lim_(x -> pi/4) 1/(cosx + sinx)    ...[(because x -> pi/4","  x ≠ pi/4),(therefore cos x ≠sinx),(therefore x - sin x ≠0)]`

= `(lim_(x -> pi/4) (1))/(lim_(x -> pi/4) (cosx + sinx))`

= `1/(cos(pi/4) + sin(pi/4))`

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

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

= `1/sqrt(2)`.

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अध्याय 7: Limits - Exercise 7.4 [पृष्ठ १४८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Exercise 7.4 | Q II. (3) | पृष्ठ १४८

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