Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By Plato

Math Help - lim (1+1/x)^x

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    Croatia
    Posts
    24

    lim (1+1/x)^x

    I need a proof that lim(1+1/x)^x=e.
    We defined that e=lim(1+1/n)^n and I need to prove that it's same for x (real number).
    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Bradyns's Avatar
    Joined
    Apr 2013
    From
    Maitland, NSW, Australia
    Posts
    39
    Thanks
    12

    Re: lim (1+1/x)^x

    The question seems trivial, as:

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

    Is the exact same statement as:

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

    Just let n = x.
    n \epsilon \mathbb{R}
    x \epsilon \mathbb{R}

    Though, this also seems trivial.. Even if they were integers or natural numbers, it would still approach the value of 'e' as the limit went to infinity.

    Have you done the proof for the 'n' in class?
    Last edited by Bradyns; April 21st 2013 at 05:43 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: lim (1+1/x)^x

    Quote Originally Posted by Bradyns View Post
    The question seems trivial, as:
    \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e
    Is the exact same statement as:
    \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}=e
    Just let n = x.
    n \epsilon \mathbb{R}
    x \epsilon \mathbb{R}
    Though, this also seems trivial.. Even if they were integers or natural numbers, it would still approach the value of 'e' as the limit went to infinity.
    That is no proof. Real numbers are not all integers.

    To do this we need to have two additional facts: assume that x\ge 1 then
    \\1)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n+1} \right )^{n}=e\\\\2)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n+1}=e

    Now use the floor function (greatest integer function): \left\lfloor x \right\rfloor  \leqslant x < \left\lfloor x \right\rfloor  + 1.

    Thus \left( {1 + \frac{1}{{\left\lfloor x \right\rfloor }}} \right) \geqslant \left( {1 + \frac{1}{x}} \right) \geqslant \left( {1 + \frac{1}{{\left\lfloor x \right\rfloor  + 1}}} \right)

    So {\left( {1 + \frac{1}{{\left\lfloor x \right\rfloor }}} \right)^{\left\lfloor x \right\rfloor  + 1}} \geqslant {\left( {1 + \frac{1}{x}} \right)^x} \geqslant {\left( {1 + \frac{1}{{\left\lfloor x \right\rfloor  + 1}}} \right)^{\left\lfloor x \right\rfloor }}
    Thanks from Gusbob
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: lim (1+1/x)^x

    We know that:

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

    \text{Let } N = \frac{1}{n}

    and as n \rightarrow \infty so N \rightarrow 0

    \therefore  \lim_{n\rightarrow \infty }  e^{\ln{\left( 1+\frac{1}{n} \right)^{n}}} = \lim_{N\rightarrow 0 } e^{\ln{\left( 1+N \right)^{\frac{1}{N}}}}


    \begin{align*}\lim_{N \rightarrow 0 } e^{\ln{(1+N)^{1/N}}} =& \lim_{N \rightarrow 0 } e^{\frac{\ln{(1+N)}}{N}}\\=& e^{\lim_{N \rightarrow 0} \frac{\ln{(1+N)}}{N}}.......[\text{by power rule of limit}]\end{align*}

    Using I'Hopital's rule:

    \lim_{N \rightarrow 0} \frac{\ln{(1+N)}}{N} = \lim_{N \rightarrow 0} \frac{\frac{1}{1+N}}{1} = 1

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


    \text{Q.E.D}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: lim (1+1/x)^x

    Quote Originally Posted by x3bnm View Post
    \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n} = e^{1} = e

    @x3bnm, Did you read the OP?
    Had you, you would have noted that the fact that \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e was never in question.
    In fact, the OP stipulates that is the definition of e.

    The you are asked to show that {\lim _{x \to \infty }}{\left( {1 + \frac{1}{x}} \right)^x} = e.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: lim (1+1/x)^x

    Quote Originally Posted by Plato View Post
    @x3bnm, Did you read the OP?
    Had you, you would have noted that the fact that \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e was never in question.
    In fact, the OP stipulates that is the definition of e.

    The you are asked to show that {\lim _{x \to \infty }}{\left( {1 + \frac{1}{x}} \right)^x} = e.
    Sorry Plato. My bad. I thought(erroneously) he wanted to know why it's equal to e. Readers...please disregard my last post. Sorry.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum