Results 1 to 6 of 6

Math Help - Limit of e

  1. #1
    Newbie
    Joined
    Sep 2006
    Posts
    16

    Limit of e

    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by JimmyT View Post
    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
    Hi,
    this one looks like a trick - but it works:

    \left(1+\frac{2}{n}\right)^n = \left(1+\frac{1}{\frac{n}{2}}\right)^n =
    \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r  ight)^2

    Now substitute n/2 = y. Then you get:


    \left( \left(1+\frac{1}{y}\right)^y\right)^2

    \lim_{y\rightarrow \infty}{\left(1+\frac{1}{y}\right)^y}=e

    Therefore \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r  ight)^2=e^2

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by JimmyT View Post
    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
    If you want full rigour you will have to fill in the detail yourself but:

    Put m=n/2, then

    f(n)=[1+(2/n)]^n=[1+m]^(2m) = {[1+m]^m}^2

    so

    lim(n->infty) [1+(2/n)]^n = lim(m->infty) {[1+m]^m}^2
    ...................................= {lim(m->infty) [1+m]^m}^2=e^2.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2006
    Posts
    16
    Along the same lines, how would you manipulate the lim (n to inf) of
    (3n)^(1/(2n)) to get 1?

    Jim
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by CaptainBlack View Post
    If you want full rigour you will have to fill in the detail yourself but:
    Actually what Earboth did was rigorous. He used the limit composition rule for sequences.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by JimmyT View Post
    Along the same lines, how would you manipulate the lim (n to inf) of
    (3n)^(1/(2n)) to get 1?

    Jim
    I am not going to write infinity because all limits of sequences are taken long infinite.

    You have,
    \lim (3n)^{\frac{1}{2n}}=\lim 3^{\frac{1}{2n}}n^{\frac{1}{2n}}=\lim \sqrt{3}^{1/n}\cdot \sqrt{n}^{1/n}
    We divide the problem into two limit problems,

    1) \lim \sqrt{3}^{1/n}=\sqrt{3}^0=1 (sequence composition rule).

    2) 1\leq \sqrt{n}^{1/n}\leq n^{1/n}
    Squeeze theorem, the limit,
    \lim n^{1/n}=1 is well known.

    Thus the answer is 1\cdot 1 =1
    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