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L'Hospital's rule

  1. Apr 14, 2003 #1
    Q1) lim (cosx)(1/x^2)
    x-->0

    I tried to find

    lim (1/x^2) ln cos x
    x-->0

    but got stuck after differentiate it several times.

    Thanks in advance
     
  2. jcsd
  3. Apr 14, 2003 #2
    L'Hospital is quite complicated to be applied here...

    (cosx)^(1/x^2)=[1-(sinx)^2]^(1/(2*(x^2)))=

    ={[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))

    lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))=
    x->0

    =lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^lim(-((sinx)^2)/(2*(x^2)))=
    x->0____________________________x->0
    =e^(-1/2)=1/sqrt(e);

    I used sinx/x -> 1 as x->0 and

    lim (1+Xn)^(1/Xn)=e, where
    x->0

    lim Xn = 0, Xn >0 or Xn <0...
    n->infinity

    I must remember you L'Hospital's rule :

    lim f(x)/g(x)=lim f'(x)/g'(x),
    x->a_______x->a

    and
    lim f(x)=0 or infinity
    x->a
    lim g(x)=0 or infinity
    x->a
    ...
     
    Last edited: Apr 14, 2003
  4. Apr 14, 2003 #3
    Thank you bogdan!

    I just got the answer before viewing your thread. I made a very silly mistake when doing this question at the very beginning.

    Here is my approach.
    lim (cosx)(1/x2)
    x-->0

    Let y=(cosx)(1/x2)
    take natural log on both sides
    ln y = (1/x2) ln cos x

    Lim (1/x2) ln cos x
    x-->0

    =Lim ln(cosx)/x2 (which is an indeterminate form of 0/0)
    x-->0

    =Lim -sinx/(2xcosx)
    x-->0

    =Lim -(sinx)/x * 1/(2cosx)
    x-->0

    =-1/2

    Therefore
    lim (cosx)(1/x^2) = e-1/2
    x-->0
     
  5. Apr 14, 2003 #4
    Very nice...and simple...
     
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