Results 1 to 13 of 13

Math Help - the limit of (1+1/n)^n=e

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    10

    the limit of (1+1/n)^n=e

    hallo,

    does any one knows how to prove that lim(1+1/n)^n=e when n goes to zero.

    thanks

    omri
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by omrimalek View Post
    hallo,

    does any one knows how to prove that lim(1+1/n)^n=e when n goes to zero.

    thanks

    omri
    I assume that you made a typo because \lim_{n \mapsto 0}\left(1+\frac1n\right)^n = 1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hi

    L=\lim_{n \to 0} \left(1+\frac 1n\right)^n

    \ln(L)=\lim_{n \to 0} n \cdot \ln \left(1+\frac 1n\right)

    But when n \to 0 ~,~ \frac 1n \gg 1

    Therefore \ln(L)=\lim_{n \to 0} n \cdot \ln \left(\frac 1n\right)

    Substituting u=\frac 1n, we get :

    \ln(L)=\lim_{u \to \infty} \frac 1u \cdot \ln(u)=0.

    Therefore, L = 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth View Post
    I assume that you made a typo because \lim_{n \mapsto 0}\left(1+\frac1n\right)^n = 1
    May be he means as n \to \infty ??

    Which can be done by taking logs then showing the limit is 1 by any number of methods including L'Hopitals rule.

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2008
    Posts
    10

    sorry i made a mistake...

    i ment when n goes to infinity...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e

    This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
    Last edited by wingless; June 28th 2008 at 02:07 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wingless View Post
    \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e

    This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
    It can be proved but that requires assuming some other definition which would otherwise be proved from this definition .....

    What can be done without assuming any definitions is proving that the limit exists and that it lies between 2 and 3.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2008
    Posts
    10

    thank you all!

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Let \varphi = \lim_{x \to \infty } \left( {1 + \frac{1}<br />
{x}} \right)^x . (Assuming the limit does exist.)

    Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

    \ln \varphi = \ln \Bigg[ {\lim_{x \to \infty } \left( {1 + \frac{1}<br />
{x}} \right)^x } \Bigg] = \lim_{x \to \infty } x\ln \left( {1 + \frac{1}<br />
{x}} \right).

    Make the substitution u=\frac1x,

    \ln \varphi =\lim_{u\to0}\frac1u\ln(1+u). Since \ln (1 + u) = \int_1^{1 + u} {\frac{1}<br />
{\tau }\,d\tau } ,\,1 \le\tau\le 1 + u,

    \frac{1}<br />
{{1 + u}} \le \frac{1}<br />
{\tau } \le 1\,\therefore \,\frac{u}<br />
{{1 + u}} \le \int_1^{1 + u} {\frac{1}<br />
{\tau}\,d\tau} \le u. So \frac1{1+u}\le\frac1u\ln(1+u)\le1.

    Take the limit when u\to0, then by the Squeeze Theorem we can conclude that \lim_{u\to0}\frac1u\ln(1+u)=1.

    Finally \ln\varphi=1\,\therefore\,\varphi=e.\quad\blacksqu  are
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Moo View Post
    Hi
    You just have to be careful about one thing over here. You assumed the limit exists. How do you know the limit exists? Note, this is why Krizalid wrote "assuming it exists". To prove this we can note the function 0,\infty)\mapsto \mathbb{R}" alt="f0,\infty)\mapsto \mathbb{R}" /> defined as f(x) = (1+\tfrac{1}{x})^x is increasing since f' >0. Therefore, the sequence x_n = (1+\tfrac{1}{n})^n is an increasing sequence. Now show that \{ x_n\} is bounded. So we have a increasing bounded sequence and therefore we have a limit.

    Quote Originally Posted by wingless View Post
    \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e

    This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
    That is not how I like to define e. I like to define \log x = \smallint_1^x \tfrac{d\mu}{\mu}. And we can define e to be the (unique) number such that \log (e) = 1. If we define e this way then it would follow that (1+\tfrac{1}{n})^n\to e. But whatever, it depends on your style of defining logarithmic functions. I just find that the approach I use it the cleanest and smoothest.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by wingless View Post
    \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e

    This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
    It is not the definition but a definition, how one proves it depends on what one is supposed to know, and how it has been defined. It is quite common for it to be defined as the base of natural logarithms.

    RonL
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by CaptainBlack View Post
    It is not the definition but a definition, how one proves it depends on what one is supposed to know, and how it has been defined. It is quite common for it to be defined as the base of natural logarithms.

    RonL
    Thanks to you both, TPH and CaptainBlack. I know that there are tons of definitions for e, but this definition is the oldest one so it seemed me nonsense to prove it.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I will present this proof for what it's worth. Though, the posters proof builds

    from this. This proof builds on the differentiability of ln(x). To be more exact

    the

    derivative of ln(x) at x=1.

    Using the definition of a derivative, since 1/x=1, we get:

    1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}

    =\lim_{h\to 0}\frac{ln(1+h)}{h}

    =\lim_{h\to 0} ln(1+h)^{\frac{1}{h}}

    Therefore, it follows that:

    e=e^{\lim_{h\to 0} ln(1+h)^{\frac{1}{h}}}

    from the continuity of e^{x} can be written this way:

    e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}

    Now, we recognize this limit. The others play off it.

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

    merely let t=\frac{1}{x}.

    This changes the limit to x\to 0 and we have said limit.
    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