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

Math Help - Limit Proof

  1. #1
    Junior Member
    Joined
    Nov 2005
    Posts
    31

    Limit Proof

    Hey, I am trying to prove that:

    lim(nth root of n)=1 as n approaches infinity and that

    lim(nth root of n+1)=1 as n approaces infinity

    I do not know how to go about it. Can anybody help??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Is using the delta-epsilon definition of a limit or a less rigorous method?

    Start with considering any function f(n)=n^\left(\frac{1}{n}\right). You can do this both graphically and analytically. For the first way, use a calculator to see the trend as n tends to infinity. For the second way, look at the exponential part of the function. What is \lim_{n\rightarrow\infty}\frac{1}{n}? And what is any number or variable to that answer?
    Last edited by Jameson; January 6th 2006 at 04:34 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    This happens to be a famous limit
    I will prove that \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1.
    Notice that the function n^{\frac{1}{n}} is a monotonic decreasing function, which has a lower bound. Thus, \lim_{n\rightarrow \infty} n^{\frac{1}{n}} has a limit. Call that L thus,
    L= \lim_{n\rightarrow \infty} n^{\frac{1}{n}}
    thus,
    \ln L=\ln (\lim_{n\rightarrow \infty} n^{\frac{1}{n}})
    But because \ln is countinous for that interval,
    \ln L=\lim_{n\rightarrow \infty} \ln (n^{\frac{1}{n}})
    Thus,
    \ln L=\lim_{n\rightarrow \infty} \frac{\ln n}{n}
    But this fits the necessary condition of L'Hopital's Rule,
    \ln L=\lim_{n\rightarrow \infty} \frac{1}{n}=0
    Thus, \ln L=0 thus, L=1
    Finally putting all of this together,
    \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Jameson
    Is using the delta-epsilon definition of a limit or a less rigorous method?

    Start with considering any function f(n)=n^\left(\frac{1}{n}\right). You can do this both graphically and analytically. For the first way, use a calculator to see the trend as n tends to infinity. For the second way, look at the exponential part of the function. What is \lim_{n\rightarrow\infty}\frac{1}{n}? And what is any number or variable to that answer?
    Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    For the second way, look at the exponential part of the function. What is \lim_{n\rightarrow\infty}\frac{1}{n}? And what is any number or variable to that answer?
    This is invalid as it pays no attention to the rate of growth of what is
    being raised to 1/n. Such an argument would apply to

    \lim_{n\rightarrow\infty}f(n)^{\frac{1}{n}}

    for any real function f(n).

    In particular it would appy when f(n)=n^n.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.
    - its actualy false (fallacious?), see my other post

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by TexasGirl
    Hey, I am trying to prove that:

    lim(nth root of n)=1 as n approaches infinity and that

    lim(nth root of n+1)=1 as n approaces infinity

    I do not know how to go about it. Can anybody help??
    Now:

    n^{\frac{1}{n}}=e^{\frac{ \ln(n)}{n}},

    and as the exponential function is continuous if the limit exists:

    \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=e^{\lim_{n\rightarrow \infty}\frac{ \ln(n)}{n}},

    so we need only worry about:

    \lim_{n\rightarrow \infty}\frac{ \ln(n)}{n}.

    Now this is dealt with elsewhere in this thread by ThePerfectHacker, but
    an informal short cut is available which tells us what this limit is.

    The sort cut is that log functions grow more slowly than any power
    of its argument, that is for all \alpha \epsilon \mathbb{R}, there exists a k such that:

    \ln (x) < k.x^{\alpha},

    for all x sufficiently large. Which is sufficient to show that

    \lim_{n\rightarrow \infty}\frac{ \ln(n)}{n^{\beta}}=0,

    for any \beta \epsilon \mathbb{R}. Of course this can also be demonstrated
    using L'Hopital's rule as is done for \beta = 1 elsewhere in this thread.

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2005
    Posts
    31

    Many Thanks

    You guys are awesome. I don't know what I would do without you. Thanks a million!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by CaptainBlack
    - its actualy false (fallacious?), see my other post

    RonL
    What are you talking about CaptainBlack.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    What are you talking about CaptainBlack.
    Jameson's statement is a fallacy - you need to consider the rate of
    growth of what is raised to 1/n not just that the limit
    of the exponent is 0. Its explained in my response to
    Jameson's post.

    RonL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by CaptainBlack
    Jameson's statement is a fallacy - you need to consider the rate of
    growth of what is raised to 1/n not just that the limit
    of the exponent is 0. Its explained in my response to
    Jameson's post.

    RonL
    I know, I told him that.

    For the second thing you cannot just take the natural logarithm of the limit you must first prove it exists. It is a common mistake by just taking the natural logarithm because we are assuming it exits.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by CaptainBlack
    This is invalid as it pays no attention to the rate of growth of what is
    being raised to 1/n. Such an argument would apply to

    \lim_{n\rightarrow\infty}f(n)^{\frac{1}{n}}

    for any real function f(n).

    In particular it would appy when f(n)=n^n.

    RonL
    I don't understand your point. Obviously lim_{x\rightarrow\infty}n^n=\infty I was simply telling her to analytically think about the trend of the exponent. If I made a mistake I apologize, but I really don't see how I was false in showing that as n approaches infinity, the function becomes n^0
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by ThePerfectHacker
    Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.
    Yes it did lack rigor. I like your method better. I make a quick post and I as I said above I was trying to analytically show the trend of the function. I'm still not seeing my fallacy though.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    I don't understand your point. Obviously lim_{x\rightarrow\infty}n^n=\infty I was simply telling her to analytically think about the trend of the exponent. If I made a mistake I apologize, but I really don't see how I was false in showing that as n approaches infinity, the function becomes n^0
    Your argument about the exponent if valid would also prove that:

    \lim_{x\rightarrow\infty}(n^n)^{1/n}=1,

    as here also the outer exponent goes to zero and as we know anything
    raised to the power zero (if non-zero) is unity.

    RonL
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Jameson, your problem, informally was that you looked at the exponent you should have also seen the base thus it is of the form \infty^0 no conclusion could be drawn.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Proof of limit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 14th 2011, 12:37 PM
  2. [SOLVED] limit proof
    Posted in the Differential Geometry Forum
    Replies: 18
    Last Post: December 23rd 2010, 08:48 PM
  3. Another limit proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 11th 2008, 11:38 AM
  4. limit proof
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 20th 2008, 05:22 PM
  5. Limit Proof
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 9th 2007, 12:28 PM

Search Tags


/mathhelpforum @mathhelpforum