Results 1 to 12 of 12

Math Help - Prove a limit

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    94
    Prove:

    Limit as n --> infinity

    n/(n^2+3n+1) -->0

    I just don't know how to start here. What algebraic property can I use to get started?

    Will this work?

    |n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?

    Thanks.
    Last edited by mr fantastic; September 12th 2009 at 05:58 PM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1490
    Quote Originally Posted by cgiulz View Post
    Prove:

    Limit as n --> infinity

    n/(n^2+3n+1) -->0

    I just don't know how to start here. What algebraic property can I use to get started?

    Thanks.
    This integral is of the indeterminate form

    \frac{\infty}{\infty}

    so L'Hospital's Rule can be used.


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

     = \lim_{x \to \infty}\frac{1}{2n + 3}

     \to \frac{1}{\infty}

     \to 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by cgiulz View Post
    Prove:

    Limit as n --> infinity

    n/(n^2+3n+1) -->0

    I just don't know how to start here. What algebraic property can I use to get started?

    Will this work?

    |n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?

    Thanks.
    The usual thing to do is \frac{n}{n^2 + 3n + 1} = \frac{1}{n + 3 + \frac{1}{n}} \rightarrow 0 as n \rightarrow \infty.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2009
    Posts
    94
    Thanks for the prompt responses.

    Is my method invalid? Or, just unorthodox?

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    of course it's valid too, you're just using the definition to prove that the limit is in efect, zero.
    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 cgiulz View Post
    Thanks for the prompt responses.

    Is my method invalid? Or, just unorthodox?

    Thanks again
    To apply your method properly you would need to set up the sort of inequality required by the pinching theorem (also known as the sandwich theorem or squeeze theorem).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by mr fantastic View Post
    The usual thing to do is \frac{n}{n^2 + 3n + 1} = \frac{1}{n + 3 + \frac{1}{n}} \rightarrow 0 as n \rightarrow \infty.
    would not be better if you divided by n^2 instead n??
    Follow Math Help Forum on Facebook and Google+

  8. #8
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    I assume from your title, and the fact that this thread is posted in a pretty advanced sub-forum, that you wish to formally prove the statement

    \lim_{n\to\infty}\frac{n}{n^2+3n+1}.

    To do this, the epsilon-delta definition is in order.

    The statement \lim_{x\to\infty}f(x)=L means that for each \epsilon>0, there exists a [MAth]M>0[/tex] such that |f(x)-L|<\epsilon whenever x>M.

    In your case, it is not difficult to see that for any \epsilon>0 a number M can be found such that

    |f(n)-0|<\epsilon whenever n>M

    This implies that the limit is ,in fact, 0.

    For the proof of n\to{-\infty}, just take values of n which are less than some number N<0.
    Last edited by VonNemo19; September 13th 2009 at 08:37 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2009
    Posts
    94
    Yes, I do and thanks. I was primarily looking for a way to get started, and I can fill in all the formalities. This is my first analysis class, so all responses are extremely useful, and somewhat ease the frustration of taking such a tedious class.

    Thanks everyone!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,987
    Thanks
    1650
    Quote Originally Posted by mr fantastic View Post
    To apply your method properly you would need to set up the sort of inequality required by the pinching theorem (also known as the sandwich theorem or squeeze theorem).
    Since you had already proved that |n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n all you need to do is observe that, since n is a positive integer, 0< |n/(n^2 +3n + 1)| < 1/n and both 0 and 1/n go to 0 as n goes to infinity.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,987
    Thanks
    1650
    Quote Originally Posted by xalk View Post
    would not be better if you divided by n^2 instead n??
    Good heavens, no! then you would have both numerator and denominator going to 0 and "0/0" is as indeterminate as " \infty/\infty".
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by HallsofIvy View Post
    Good heavens, no! then you would have both numerator and denominator going to 0 and "0/0" is as indeterminate as " \infty/\infty".
    (Pssst ..... Actually the denominator would go to 1 ....).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  2. How to prove? (limit)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 22nd 2009, 09:12 AM
  3. how to prove that any limit ordinal is really a limit?
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: August 9th 2009, 01:25 AM
  4. Using the Central Limit Theorem to prove a limit
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 23rd 2009, 11:09 AM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum