मराठी

Limx→π4tan3x-tanxcos(x+π4) - Mathematics

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प्रश्न

`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`

बेरीज
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उत्तर

Given, `lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`

= `lim_(x -> pi/4) (tanx(tan^2x - 1))/(cos(x + pi/4))`

= `lim_(x -> pi/4) (tanx (tan^2x - 1))/(cos(x + pi/4))`

= `lim_(x -> pi/4) tan x * lim_(x -> pi/4) [(-(1 - tan^2x))/(cos(x + pi/4))]`

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

= `lim_(x -> pi/4) - (1 + tan x) * lim_(x -> pi/4) ((1 - tanx)/(cos(x + pi/4)))`

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

= `-2 xx lim_(x -> pi/4) (sqrt(2) (1/sqrt(2) cos x - 1/sqrt(2) sinx))/(cos x * cos (x + pi/4))`

= `-2sqrt(2) lim_(x -> pi/4) ([cos  pi/4 * cos x - sin  pi/4 sin x])/(cosx * cos(x + pi/4))`

= `lim_(x -> pi/4) (-2sqrt(2) * cos(x + pi/4))/(cosx * cos(x + pi/4))`

= `(-2sqrt(2))/(cos  pi/4)`  ....(Taking limit)

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

= `-2 xx 2`

= – 4.

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

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 49 | पृष्ठ २४१

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