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Question Number 102020 by Dwaipayan Shikari last updated on 06/Jul/20

lim_(x→0) (tan((π/4)−x))^(1/x)

limx0(tan(π4x))1x

Answered by john santu last updated on 06/Jul/20

lim_(x→0) (1+(tan ((π/4)−x)−1))^(1/x)   =e^(lim_(x→0) (((tan ((π/4)−x)−1)/x)))   =e^(lim_(x→0) ((−sec^2  ((π/4)−x))/1)) = e^(−2)  = (1/e^2 )  (JS ⊛)

limx0(1+(tan(π4x)1))1x=elimx0(tan(π4x)1x)=elimx0sec2(π4x)1=e2=1e2(JS)

Answered by Ar Brandon last updated on 06/Jul/20

Let y=lim_(x→0) [tan((π/4)−x)]^(1/x) ⇒lny=lim_(x→0) {((ln[tan((π/4)−x)])/x)}  ⇒lny=lim_(x→0) {−sec((π/4)−x)cosec((π/4)−x)}               =−sec((π/4))cosec((π/4))=−2  ⇒lim_(x→0) [tan((π/4)−x)]^(1/x) =(1/e^2 )

Lety=limx0[tan(π4x)]1xlny=limx0{ln[tan(π4x)]x}lny=limx0{sec(π4x)cosec(π4x)}=sec(π4)cosec(π4)=2limx0[tan(π4x)]1x=1e2

Commented by Dwaipayan Shikari last updated on 06/Jul/20

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