Results 1 to 3 of 3

Math Help - A limit

  1. #1
    Banned
    Joined
    Jul 2009
    Posts
    107

    A limit

    Assuming that :

    \lim_{x\to a}f(x) = l\neq 0,and

    \lim_{x\to a}g(x)= m>0 ,then

    Prove that:

    \lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}

    by using the ε-δ definition of a limit:

    So we must prove that:

    Given an ε>0 we must find a δ>0 such that:

    for all ,x : 0<|x-a|<δ \Longrightarrow | f(x)^{  g(x) } -l^m|<\epsilon.

    The point now is how do we manipulate ,  |f(x)^{g(x)}-l^m|
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    we may use
    f(x)^{g(x)}=e^{g(x)\ln f(x)}

    |e^{g(x)\ln f(x)}-l^m|=|l^m||\frac{e^{g(x)\ln f(x)}}{l^m}-1|=

    =|l^m||\frac{e^{(m+dp)\ln {(l+dq)}}}{l^m}-1|=

    =|l^m||\frac{e^{(m+dp)\ln {l(1+ \frac{dq}{l})}}}{l^m}-1|=

    =|l^m||\frac{   e^{  (m+dp)( \ln l+ \frac{dq}{l})  }   }{ l^m }-1|=

    =|l^m||\frac{   e^{   m \ln l +dp \ln l + \frac{mdq}{l}  } }{ l^m }-1|=

    =|l^m||\frac{  l^m e^{    { dp \ln l + \frac{mdq}{l}}  } }{ l^m }-1|=

    =|l^m|| { {dp \ln l} + \frac{mdq}{ l }}  | <

    <  |l^m| (  |{dp \ln l}| +| \frac{mdq}{ l }|) < | dd | |l^m|  (| \ln l | + |\frac{m}{l}   |) < \varepsilon

    | dd | < \frac{\varepsilon}{ |l^m|(| \ln l | + |\frac{m}{l}   |) }
    where
    <br />
dd=max(dp,dq)<br />
    Last edited by zzzoak; April 21st 2010 at 03:01 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Jul 2009
    Posts
    107
    Quote Originally Posted by zzzoak View Post
    we may use
    f(x)^{g(x)}=e^{g(x)\ln f(x)}

    |e^{g(x)\ln f(x)}-l^m|=|l^m||\frac{e^{g(x)\ln f(x)}}{l^m}-1|=

    =|l^m||\frac{e^{(m+dp)\ln {(l+dq)}}}{l^m}-1|=

    =|l^m||\frac{e^{(m+dp)\ln {l(1+ \frac{dq}{l})}}}{l^m}-1|=

    =|l^m||\frac{   e^{  (m+dp)( \ln l+ \frac{dq}{l})  }   }{ l^m }-1|=

    =|l^m||\frac{   e^{   m \ln l +dp \ln l + \frac{mdq}{l}  } }{ l^m }-1|=

    =|l^m||\frac{  l^m e^{    { dp \ln l + \frac{mdq}{l}}  } }{ l^m }-1|=

    =|l^m|| { {dp \ln l} + \frac{mdq}{ l }}  | <

    <  |l^m| (  |{dp \ln l}| +| \frac{mdq}{ l }|) < | dd | |l^m|  (| \ln l | + |\frac{m}{l}   |) < \varepsilon

    | dd | < \frac{\varepsilon}{ |l^m|(| \ln l | + |\frac{m}{l}   |) }
    where
    <br />
dd=max(dp,dq)<br />
    Thanks ,but i am completely lost.

    Where do you get dp,dq ??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  2. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  3. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum