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