Menu Close

Find-the-limit-x-0-lim-tan-x-x-1-x-2-




Question Number 82494 by niroj last updated on 21/Feb/20
      Find the limit    _(x→0) ^(lim)  (((tan x)/x))^(1/x^( 2) )
Findthelimitlimx0(tanxx)1x2
Commented by abdomathmax last updated on 21/Feb/20
let f(x)=(((tanx)/x))^(1/x^2 )  ⇒f(x)=e^((1/x^2 )ln(((tanx)/x)))   u(x)=tanx =u(0)+(x/(1!))u^((1)) (o) +(x^2 /(2!))u^((2)) (0)+(x^3 /(3!))u^((3)) (0)+...  u^′ (x)=1+tan^2 x ⇒u^′ (0)=1  u^((2)) (x)=2tanx(1+tan^2 x) ⇒u^((2)) (0)=0  u^((3)) (x) =2(1+tan^2 x)(1+tan^2 x)+  +(2tanx)(2tanx(1+tan^2 x)) ⇒u^((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)
letf(x)=(tanxx)1x2f(x)=e1x2ln(tanxx)u(x)=tanx=u(0)+x1!u(1)(o)+x22!u(2)(0)+x33!u(3)(0)+u(x)=1+tan2xu(0)=1u(2)(x)=2tanx(1+tan2x)u(2)(0)=0u(3)(x)=2(1+tan2x)(1+tan2x)++(2tanx)(2tanx(1+tan2x))u(3)(0)=2tanx=x+x33+o(x5)tanxx=1+x23+o(x4)ln(tanxx)=ln(1+x23+o(x4))x231x2ln(tanxx)13limx0f(x)=e13=3e
Commented by jagoll last updated on 22/Feb/20
lim_(x→0)  (1+((tan x)/x)−1)^(1/x^2 )  =   lim_(x→0)  (1+((tan x−x)/x))^(1/x^2 )  =  e^(lim_(x→0)  (((tan x−x)/x^3 ))) = e^(lim_(x→0)  (((sec^2 x−1)/(3x^2 ))))   = e^(lim_(x→0)  (((1−cos^2 x)/(3x^2  cos^2 x))))  = e^(1/3)
limx0(1+tanxx1)1x2=limx0(1+tanxxx)1x2=elimx0(tanxxx3)=elimx0(sec2x13x2)=elimx0(1cos2x3x2cos2x)=e13
Commented by john santu last updated on 22/Feb/20
nice
nice
Commented by niroj last updated on 22/Feb/20
 great.
great.

Leave a Reply

Your email address will not be published. Required fields are marked *