Results 1 to 10 of 10

Math Help - limit = e

  1. #1
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182

    limit = e

    Can someone show me how:

    \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by Greengoblin View Post
    Can someone show me how:

    \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e
    Say that this limit exists and it equals A.

    Then =\frac{1}{t}\ln(1+t)=\ln(A)

    The LHS is also \frac{\ln(1+t)}{t} and when evaluating at t=0 you get 0/0, which is indeterminate and L'Hopitals can be used.

    So \ln(A)=\frac{\frac{1}{1+t}}{1}

    Try the limit again, and use algebra to solve for A again.

    edit: This is in the precalc section, so forgive me for using calculus.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by Greengoblin View Post
    Can someone show me how:

    \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e
    Another way is to sub u=\frac{1}{t} so that u \to \infty~\text{as}~t \to 0 to get the famous limit:

    \lim_{u \to \infty} \left(1+\frac{1}{u}\right)^u = e

    @Jameson: I am sure that Greengoblin's level is beyond precal.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Yes that would appear so after looking some more
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182
    Thank you both so much, I understand it now.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Alternate solution:

    recall that e^{\ln X} = X for any X > 0.

    so that:

    \lim_{t \to 0}(1 + t)^{\frac 1t} = \lim_{t \to 0}\text{exp} \left( \frac 1t \ln (1 + t) \right)

    = \text{exp} \left( \lim_{t \to 0} \frac {\ln (1 + t)}t \right)

    now since \lim_{t \to 0} \frac {\ln (1 + t)}t = 1 (By L'Hopital's rule, or power series or ...)

    we have that the limit is \text{exp}(1) = e


    This is very similar to Jameson's method, both of these methods you should consider when taking the limit of an expression with the variable in the power.

    That is, let A and B be expressions in a single variable in which the variable appears, then we can usually find various limits that give indeterminate forms by doing the following:

    The method Jameson used:

    Say y = A^B

    Then \ln y = B \ln A

    so that \lim \ln y = \lim B \ln A

    and so \lim y = e^{\lim B \ln A} by taking the anti-log of both sides.


    The method I used:

    simply write y = A^B as e^{\ln A^B} = e^{B \ln A}, and take the limit of both sides.

    note that \lim e^X = e^{\lim X}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182
    Thanks, but could you explain how \lim e^X=e^{\lim X} ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2008
    Posts
    792
    If \lim_{x \to a} f(x) = L and g is continuous on L, then \lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a} f(x)\right) = g(L)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Greengoblin View Post
    Thanks, but could you explain how \lim e^X=e^{\lim X} ?
    the rigorous theory might get a bit complicated, but it is somewhat similar to the fact that we can factor constants out side of limits. since the limit deals with X and not e, taking the limit of e to some power really only affects the power, since anything that is not a variable in X is not affected by the limit.

    EDIT: Chop Suey was a bit more rigorous .
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member Greengoblin's Avatar
    Joined
    Feb 2008
    From
    UK
    Posts
    182
    Thanks, I got it now. I need to have a good look at limits again, as I haven't looked at them for ages and have forgotten alot.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: August 26th 2010, 10:59 AM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  4. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  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