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Evaluate the following : limx→π4[(sinx-cosx)22-sinx-cosx] - Mathematics and Statistics

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प्रश्न

Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`

योग
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उत्तर

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`

= `lim_(x -> pi/4) (1 - sin2x)/(sqrt(2) - (sin x + cos x))`

= `lim_(x -> pi/4) (1 - sin2x)/(sqrt(2) - sqrt(1 + sin2))`

Put 1 + sin 2x = t

∴ sin 2x = t – 1

As `x -> pi/4, "t" -> 1 + sin 2 (pi/4)`

∴ `"t" -> 1 + sin  pi/2`

∴ t → 1 + 1

∴ t → 2

∴ Required limit

= `lim_("t" -> 2) (1 - ("t" - 1))/(sqrt(2) - sqrt("t"))`

= `lim_("t" -> 2) (2 - "t")/(2^(1/2) - "t"^(1/2))`

= `lim_("t" -> 2) ("t" - 2)/("t"^(1/2) - 2^(1/2))`

= `lim_("t" -> 2) 1/(("t"^(1/2) - 2^(1/2))/("t" - 2))   ...[("Divide Numerator and Denominator"),("by"  "t" - 2  "As"  "t" -> 2 ","  "t" ≠ 2),(therefore "t" - 2 ≠ 0)]`

= `1/(1/2(2)^((-1)/2)`

= `2(2)^(1/2)`

= `2sqrt(2)`

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अध्याय 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.2 | Q II. (17) | पृष्ठ १५९

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