Results 1 to 5 of 5

Math Help - How do I show that lim x->inf (1 + 1/n)^n = e ?

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    How do I show that lim x->inf (1 + 1/n)^n = e ?

    How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students - in other words, I'm bombarded with too much work at the moment to be able to learn something that is not part of my course)?

    Any input would be greatly appreciated!
    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by s3a View Post
    How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students - in other words, I'm bombarded with too much work at the moment to be able to learn something that is not part of my course)?

    Any input would be greatly appreciated!
    Thanks in advance!
    You can use logs in conjunction with L'Hopital's rule

    or you can use the binomial expansion.

    ln\left(1+\frac{1}{x}\right)^x=ln(limit)

    xln\left(1+\frac{1}{x}\right)=ln(limit)

    \frac{ln\left(1+\frac{1}{x}\right)}{\left(\frac{1}  {x}\right)}=ln(limit)

    This gives 0/0 as x goes to infinity,
    so applying L'Hopital's rule, differentiating numerator and denominator with the chain rule..

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

    =\frac{1}{\frac{x+1}{x}}=\frac{x}{x+1}

    As x approaches infinity, this approaches 1,
    hence

    1=ln(limit)\ \Rightarrow\ e^1=limit
    Follow Math Help Forum on Facebook and Google+

  3. #3
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597
    Just a quick question; is it acceptable to use the word "limit" as a variable on a test? What would you recommend?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by s3a View Post
    Just a quick question; is it acceptable to use the word "limit" as a variable on a test? What would you recommend?
    Not as a variable, no...

    I placed it in brackets for notational convenience as a shortcut,
    however, it's preferable maybe to use a capital L.

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

    then

    \lim_{x\rightarrow\infty}ln\left(1+\frac{1}{x}\rig  ht)^x=ln(L)

    \lim_{x\rightarrow\infty}xln\left(1+\frac{1}{x}\ri  ght)=ln(L)

    \lim_{x\rightarrow\infty}\frac{ln\left(1+\frac{1}{  x}\right)}{\left(\frac{1}{x}\right)}=ln(L)

    Then applying L'Hopital's rule, differentiating numerator and denominator..

    \lim_{x\rightarrow\infty}\frac{x+1}{x}=ln(L)

    1=ln(L)

    L=e

    \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right  )^x=e
    Attached Thumbnails Attached Thumbnails How do I show that lim x->inf (1 + 1/n)^n = e ?-limit.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,649
    Thanks
    1597
    Awards
    1
    Quote Originally Posted by s3a View Post
    How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students
    I am not sure what all schools teach Caculus II students.
    Here is the way I present the limit.
    We know that \frac{1}{1+n}<\ln\left(1+\frac{1}{n}\right)<\frac{  1}{n}.
    From which we get e^{\frac{n}{1+n}}<\left(1+\frac{1}{n}\right)^n<e .
    The limit is squeezed between e~\&~e.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 8th 2011, 12:05 PM
  2. show the way
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 14th 2009, 04:47 AM
  3. how to show show this proof using MAX
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 14th 2009, 12:05 PM
  4. How to show...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2008, 09:07 AM
  5. how to show -u = (-1) u ?
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 21st 2008, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum