Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - find limit by rearrangement or L'Hopital's?

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    7

    find limit by rearrangement or L'Hopital's?

    Hi, 1st post.
    According to my text book the following limit converges to  e^2 :
    \lim_{n\to\infty}(\frac {n + 3}{n + 1})^n<br />

    through rearrangement i can achieve this:

    <br />
(\frac {n * (1 + 3/n)} {n * (1 + 1/n)})^n<br />

    giving...

    <br />
\frac {(1 + 3/n)^n} {(1 + 1/n)^n}<br />

    The denominator evaluates to e but i can't see how to make the numerator equal to e^3 in order to achieve
    agreement with the textbook answer. I had a look at using L'Hopital's but
    the resulting expression after differentiation looked no simpler.

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by framer View Post
    Hi, 1st post.
    According to my text book the following limit converges to  e^2 :
    \lim_{n\to\infty}(\frac {n + 3}{n + 1})^n<br />

    through rearrangement i can achieve this:

    <br />
(\frac {n * (1 + 3/n)} {n * (1 + 1/n)})^n<br />

    giving...

    <br />
\frac {(1 + 3/n)^n} {(1 + 1/n)^n}<br />

    The denominator evaluates to e but i can't see how to make the numerator equal to e^3 in order to achieve
    agreement with the textbook answer. I had a look at using L'Hopital's but
    the resulting expression after differentiation looked no simpler.

    thanks
    Suppose the limit converges to L then

    L=\lim_{n\to\infty}(\frac {n + 3}{n + 1})^n

    taking the natural log of both sides we get

    \ln(L)=\lim_{n\to\infty} \ln [(\frac {n + 3}{n + 1})^n]

    \ln(L)=\lim_{n\to\infty} n\ln [(\frac {n + 3}{n + 1})]

    \ln(L)=\lim_{n\to\infty} \frac{\ln [(\frac {1 + 3/n}{1 + 1/n})]}{\frac{1}{n}}

    now finally using L'H rule we get after a ALOT of algebra

    \ln(L)=\lim_{n\to\infty} \frac{2n^2}{(n+3)(n+1)}

    \ln(L)=2 \iff L = e^2

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by framer View Post
    ...
    giving...

    <br />
\frac {(1 + 3/n)^n} {(1 + 1/n)^n}<br />

    The denominator evaluates to e but i can't see how to make the numerator equal to e^3 in order to achieve
    agreement with the textbook answer. ...
    You already know that

    \lim_{t \mapsto \infty}\left(1+\frac1t\right)^t = e

    Now set \frac1t = \frac3n~\implies~n = 3t

    Then

    \lim_{n \mapsto \infty}\left(1+\frac3n\right)^n~\buildrel {n=3t} \over \longrightarrow~\lim_{3t \mapsto \infty}\left(1+\frac1t\right)^{3t} = \lim_{3t \mapsto \infty}\left(\left(1+\frac1t\right)^t\right)^3 = e^3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    7
    Thanks to both of you for such prompt replies.
    Wondering if you might point out where my line of reasoning went wrong.
    I thought that if numerator has to evaluate to e^3
    then I would be satisfied if I could show that

    ((1+1/n)^n)^3  = (1+3/n)^n

    so I used the exponent identity: a^n . b^n = (a . b)^n

    to get

    ((1+1/n) . (1+1/n) . (1+1/n))^n

    = ((1 +1/n) . (1 + 2/n + 1/n^2))^n

    = (1 + 3/n + 2/n^2 + 1/n^3)^n

    != (1+ 3/n)^n<br />

    Why does this approach not work? Have I made an algebraic error?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by framer View Post
    Thanks to both of you for such prompt replies.
    Wondering if you might point out where my line of reasoning went wrong.
    I thought that if numerator has to evaluate to e^3
    then I would be satisfied if I could show that

    ((1+1/n)^n)^3  = (1+3/n)^n

    so I used the exponent identity: a^n . b^n = (a . b)^n

    to get

    ((1+1/n) . (1+1/n) . (1+1/n))^n

    ...
    You have (correctly!) noted that \frac3n = \frac1n+\frac1n+\frac1n

    That means:

    \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr  ac1n\right)

    But now you perform an algebraic salto mortale :

    \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by framer View Post
    Thanks to both of you for such prompt replies.
    Wondering if you might point out where my line of reasoning went wrong.
    I thought that if numerator has to evaluate to e^3
    then I would be satisfied if I could show that

    ((1+1/n)^n)^3 = (1+3/n)^n

    so I used the exponent identity: a^n . b^n = (a . b)^n

    to get

    ((1+1/n) . (1+1/n) . (1+1/n))^n

    = ((1 +1/n) . (1 + 2/n + 1/n^2))^n

    = (1 + 3/n + 2/n^2 + 1/n^3)^n

    != (1+ 3/n)^n<br />

    Why does this approach not work? Have I made an algebraic error?
    You're obviously correct in saying that (1 + 3/n + 2/n^2 + 1/n^3)^n != (1+ 3/n)^n.

    However, \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n IS equal to \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n since the 3/n term dominates over the 2/n^2 and 1/n^3 terms in the limit ......
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2008
    Posts
    7
    thanks for the reply, most appreciated but I have a nagging feeling of dissatisfaction with the explanation because ....

    'if'...

    \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n = e^3<br />

    wouldn't...

    \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n =

    a number just slightly larger than e^3?

    Doesn't this break the definition of a limit of a sequence?


    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by framer View Post
    thanks for the reply, most appreciated but I have a nagging feeling of dissatisfaction with the explanation because ....

    'if'...

    \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n = e^3
    " alt="

    'if'...

    \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n = e^3
    " />

    wouldn't...

    \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n =

    a number just slightly larger than e^3? Mr F says: In the limit the difference goes to zero.

    Doesn't this break the definition of a limit of a sequence?


    By the way, it doesn't actually make a difference in the limit, but it should be \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{{\color{red}3}}{n^2} + \frac{1}{n^3}\right)^n
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Apr 2008
    Posts
    7
    Sorry I somehow didn't see your post until now and now I have some questions


    Quote Originally Posted by earboth View Post
    You have (correctly!) noted that \frac3n = \frac1n+\frac1n+\frac1n

    I did? Where?

    That means:

    \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr  ac1n\right)

    But now you perform an algebraic salto mortale :

    I did!? Ok I googled Salto Mortale and it said it was Italian for deadly jump, which sounds bad!

    \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)
    My algebra is clearly crummy but could you please explain where exactly I went awry in an even more obvious way (if that is possible?)

    Is it this part of my reasoning that is wrong??:

    \left( \left( 1 + \frac{1} {n}\right) ^n \right) ^3 <br />
= \left(\left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n} \right)\right)^n ?


    I'm confused
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Well

    Quote Originally Posted by framer View Post
    Sorry I somehow didn't see your post until now and now I have some questions




    My algebra is clearly crummy but could you please explain where exactly I went awry in an even more obvious way (if that is possible?)

    Is it this part of my reasoning that is wrong??:

    \left( \left( 1 + \frac{1} {n}\right) ^n \right) ^3 <br />
= \left(\left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n} \right)\right)^n ?


    I'm confused
    IF you expand your adaptation you will see you do not get that
    Follow Math Help Forum on Facebook and Google+

  11. #11
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    Quote Originally Posted by Mathstud28 View Post
    IF you expand your adaptation you will see you do not get that
    Sure you do.

    \left[ \left(1 + \frac{1}{n}\right)^{n}\right]^{3} = \left[ \left(1 + \frac{1}{n}\right)^{3}\right]^{n} = \left[ \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right)\right]^{n}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Apr 2008
    Posts
    7
    Quote Originally Posted by o_O View Post
    Sure you do.

    \left[ \left(1 + \frac{1}{n}\right)^{n}\right]^{3} = \left[ \left(1 + \frac{1}{n}\right)^{3}\right]^{n} = \left[ \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right)\right]^{n}

    Yeah, that's the way I thought it expanded but an earlier poster
    (earboth) said that I was making a fatal error somewhere but I am still no closer to understanding where
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Apr 2008
    Posts
    7
    Quote Originally Posted by earboth View Post
    You have (correctly!) noted that \frac3n = \frac1n+\frac1n+\frac1n

    That means:

    \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr  ac1n\right)

    But now you perform an algebraic salto mortale :

    \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)
    What did earboth mean by this?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by framer View Post
    Yeah, that's the way I thought it expanded but an earlier poster
    (earboth) said that I was making a fatal error somewhere but I am still no closer to understanding where
    As you should see by now, there's a small salto but no mortale ..... earboth probably misread your working (it happens, especially late at night - I've done it myself many times)
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Apr 2008
    Posts
    7
    Ah ok, thanks Mr F. I'm not confident with my maths so I tend to distrust my own
    working which leads to confusion and negative thinking towards maths in general . I'm still having trouble understanding your argument about the last two terms are able to be ignored in the limit

    <br />
\lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{3}{n^2} + <br />
\frac{1}{n^3} \right)^n

    I'm thinking I need to review a textbook on this; do you know of any online
    content that talks about this sort of thing in depth?

    thanks
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. L'Hopital's limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2011, 06:52 PM
  2. Limit using L'H˘pital
    Posted in the Calculus Forum
    Replies: 11
    Last Post: May 25th 2010, 10:52 AM
  3. limit/l'hopital
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 14th 2009, 06:51 PM
  4. L'Hopital Limit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 25th 2009, 10:43 PM
  5. limit using l'hopital
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 24th 2007, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum