Question Number 101981 by Rohit@Thakur last updated on 05/Jul/20

lim_(x→0) (((tanx)/x))^(1/x^2 )

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

lim_(x→0) (((tanx)/x))^(1/x^2 ) =y  lim_(x→0) (1/x^2 )log(((tanx)/x))=logy  lim_(x→0) (1/x^2 )log(1+((tanx−x)/x))=logy  lim_(x→0) (1/x^2 )log(1+(((x^3 /3)+x−x)/x))=logy  lim_(x→0) (1/3) ((log(1+(x^2 /3)))/(x^2 /3))=logy  logy=(1/3)  ⇒  y=e^(1/3)

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

lim_(x→0)  (1+(((tan x)/x)−1))^(1/x^2 ) =  e^(lim_(x→0) (((tan x−x)/x)). (1/x^2 ))  = e^(lim_(x→0) (((tan x−x)/x^3 )))   =e^(lim_(x→0) ((((x+(1/3)x^3 )−x)/x^3 )))  = e^(1/3) = (e)^(1/(3  ))   (JS ⊛)

Answered by mathmax by abdo last updated on 06/Jul/20

let f(x) =(((tanx)/x))^(1/x^2 )  ⇒f(x) =e^((1/x^2 )ln(((tanx)/x)))   we have tanx =tan0 +(x/(1!))tan^′ (0) +(x^2 /(2!))tan^((2)) (0) +(x^3 /(3!))tan^3 (0) +x^5 }δ(x)  tan0 =0  tan^′ x =1+tan^2 x ⇒tan^′ (0) =1  tan^((2)) (0) =2tanx(1+tan^2 x) ⇒tan^((2)) (0) =0  tan^((3)) (0) =2(1+tan^2 x)(1+tan^2 x) +2tanx(2tanx)(1+tan^2 x) ⇒  tan^((3)) (0) =2 ⇒tanx =x +(x^3 /3) +o(x^5 ) ⇒((tanx)/x)  =1 +(x^2 /3) +o(x^4 ) ⇒  ln(((tanx)/x)) =ln(1+(x^2 /3) +o(x^4 )))∼(x^2 /3) ⇒(1/x^2 )ln(((tanx)/x)) ∼(1/3) ⇒  lim_(x→0) f(x) =e^(1/3)  =^3 (√e)