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Evaluate the following limit : limx→π[1-cosx-2sin2x] - Mathematics and Statistics

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प्रश्न

Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`

योग
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उत्तर

`lim_(x -> pi) (sqrt(1 - cosx) - sqrt(2))/(sin^2 x)`

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

= `lim_(x -> pi) ((1 - cos x) - 2)/(( 1 - cos^2x)(sqrt(1 - cos x) + sqrt(2))`

= `lim_(x -> pi) (-(1 + cos x))/((1 + cos x)(1 - cosx)(sqrt(1 - cosx) + sqrt(2))`

= `lim_(x -> pi) (-1)/((1 - cosx)(sqrt(1 - cosx) + sqrt(2)))  ...[(because x -> pi","  x ≠ pi),(therefore cos x ≠ cos pi = -1),(therefore 1 + cos x ≠ 0)]`

= `(lim_(x -> pi) (-1))/([lim_(x -> pi) (1 - cosx)] xx [lim_(x -> pi) (sqrt(1 - cosx) + sqrt(2))]`

= `(-1)/((1 - cos pi) (sqrt(1 - cos  pi) + sqrt(2))`

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

= `(-1)/(4sqrt(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 III. (2) | पृष्ठ १४८

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