Results 1 to 11 of 11

Math Help - Limits

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    94

    Cool Limits

    I have difficulties in solving problems

    1) Show that lim (1 + (1/2n))^n = sqrt e.
    n-> oo

    2)find lim (1+2/n)^n
    n-> oo

    3) lim (1 + 3/n) ^n
    n-> oo

    4) suppose 0 < Sn and Sn+1 < rSn, where r is a constant such that 0 < r < 1. Show that limn->oo Sn = 0.

    5) suppose 0 < Sn and Sn+1 >= rSn, where r is a constant such that r > 1. Show that limn->oo Sn = +oo.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by mathsohard View Post
    I have difficulties in solving problems

    1) Show that lim (1 + (1/2n))^n = sqrt e.
    n-> oo
    \displaystyle\lim_{n \to \infty}\left[1+\frac{1}{2n}\right]^n=\lim_{n \to \infty}\left(\left[1+\frac{1}{2n}\right]^{2n}\right)^{\frac{1}{2}}

    which is ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,715
    Thanks
    1642
    Awards
    1
    For #4, show that S_{n+1}<r^nS_1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    94
    Can you give me some help on showing that???
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81
    Quote Originally Posted by mathsohard View Post
    I have difficulties in solving problems

    1) Show that lim (1 + (1/2n))^n = sqrt e.
    n-> oo

    2)find lim (1+2/n)^n
    n-> oo

    3) lim (1 + 3/n) ^n
    n-> oo

    4) suppose 0 < Sn and Sn+1 < rSn, where r is a constant such that 0 < r < 1. Show that limn->oo Sn = 0.

    5) suppose 0 < Sn and Sn+1 >= rSn, where r is a constant such that r > 1. Show that limn->oo Sn = +oo.
    Pretty much for 1, 2, 3:
    \displaystyle \[\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{C}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {e^{n \cdot \ln \left( {1 + \frac{C}{n}} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } n \cdot \ln \left( {1 + \frac{C}{n}} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } n \cdot \frac{C}{n}}} = {e^C}\]

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2010
    Posts
    94
    for 2 and 3 if I wanted to use the fact (1+2/n) = (1+ 1/(n+1))(1+ 1/n) how can I do this??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    I'd have approached (2) differently:

    \displaystyle\lim_{n \to \infty}\left[1+\frac{2}{n}\right]^n=\lim_{n \to \infty}\left(\left[1+\frac{1}{\left(\frac{n}{2}\right)}\right]^{\frac{n}{2}\right)^2
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2010
    Posts
    94
    Thanks for 1, 2, 3 , I still need help on 4, 5
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by mathsohard View Post
    Thanks for 1, 2, 3 , I still need help on 4, 5
    Referring to Plato's post:

    S_1>0,\;\;0<r<1

    S_2<rS_1\Rightarrow\ S_3<rS_2\Rightarrow\ S_3<r^2S_1

    S_4<r^3S_1

    S_{n+1}<r^nS_1

    Since 0<r<1,\;\;\;\displaystyle\lim_{n \to \infty}r^n=\;?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Nov 2010
    Posts
    94
    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    0<r<1\Rightarrow\ r=\frac{a}{b}=\frac{a}{ka}

    since b>a\Rightarrow\ b=ka,\;\;k>1

    \displaystyle\Rightarrow\lim_{n \to \infty}\left(\frac{1}{k}\right)^n=0,\;\;k>1


    Or:

    \displaystyle\ 0<r<1\Rightarrow\ k=\frac{1}{r}\Rightarrow\ k>1

    \displaystyle\ r^n=\left(\frac{1}{k}\right)^n etc
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limits
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: February 10th 2010, 11:24 AM
  2. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 05:34 PM
  3. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum