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

lim_(x→0) (tan((π/4)−x))^(1/x)

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 ⊛)

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 )

Commented byDwaipayan Shikari last updated on 06/Jul/20

Great