Results 1 to 5 of 5

Math Help - limit problem

  1. #1
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64

    limit problem

    this question is the limit as t tends to infinity of T(ln(5+1/t) - ln(5))

    I think im having an algebraic problem. basicly the problem can be arranged :

    infinity times 0 =

    (ln(5+1/t) - ln(5)) / 1/T

    that makes it 0/0 so I can use L'hopitals.

    so I take the derivatives and get this :

    ((1/(5+1/T))(-1/t^2) - 1/5) / (-1/t^2)

    But im not sure what to do from here. Ive been trying different thigns but keep getting different answers. the right answers supposed to be 1/5. can someone break it down for me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294
    To start with, ln(5) is a constant. The derivative of a constant is 0. Fix that error and you should start to see things cancelling.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by frankinaround View Post
    this question is the limit as t tends to infinity of T(ln(5+1/t) - ln(5))

    I think im having an algebraic problem. basicly the problem can be arranged :

    infinity times 0 =

    (ln(5+1/t) - ln(5)) / 1/T

    that makes it 0/0 so I can use L'hopitals.

    so I take the derivatives and get this :

    ((1/(5+1/T))(-1/t^2) - 1/5) / (-1/t^2)

    But im not sure what to do from here. Ive been trying different thigns but keep getting different answers. the right answers supposed to be 1/5. can someone break it down for me?
    Take into account that for t 'large enough' is...

    (1)

    ... then multiplying (1) by t...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    Ohhhh. I always thought ln(x) was 1/x, but now I know its chain rule. So ln(5) = 1/5 times d/dx(5) so 1/5 times 0 = 0. right? really ? whoa.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by frankinaround View Post
    Ohhhh. I always thought ln(x) was 1/x, but now I know its chain rule. So ln(5) = 1/5 times d/dx(5) so 1/5 times 0 = 0. right? really ? whoa.
    Well, technically yes. But (d/dx)(constant) = 0 automatically. There is no need for the chain rule.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: August 13th 2010, 01:03 AM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. Limit problem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: June 7th 2010, 08:00 AM
  4. problem of limit
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 24th 2009, 02:48 AM
  5. 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

Search Tags


/mathhelpforum @mathhelpforum