Results 1 to 14 of 14

Thread: limits of a sequence

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    30

    limits of a sequence

    In general, how do you go about finding the limit of a sequence?

    e.g. n^{1/$\displaystyle \sqrt{n}$}

    Sorry about latex fail, its n to the power the whole of the bracket.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    Quote Originally Posted by Mathman87 View Post
    In general, how do you go about finding the limit of a sequence?

    e.g. n^{1/$\displaystyle \sqrt{n}$}

    Sorry about latex fail, its n to the power the whole of the bracket.
    What is n approaching?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    30
    It doesnt specify on the examples, but i imagine infinity?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    $\displaystyle \infty^0$ is an indeterminant form. Do you know how to solve those?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    Posts
    30
    no?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    L'Hopital's Rule
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by Mathman87 View Post
    In general, how do you go about finding the limit of a sequence?

    e.g. n^{1/$\displaystyle \sqrt{n}$}

    Sorry about latex fail, its n to the power the whole of the bracket.
    $\displaystyle \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } $

    Let,

    $\displaystyle y = n^{ \frac{1}{ \sqrt{n} } } $

    $\displaystyle ln(y) = \frac{1}{ \sqrt{n}} ln(n) $

    $\displaystyle \lim_{ n \to \infty } ln(y) = \lim_{ n \to \infty } \frac{1}{ \sqrt{n}} ln(n)$

    This is of the form $\displaystyle \frac{ \infty } { \infty } $ so we can use L'hopitals here to get

    $\displaystyle \lim_{ n \to \infty } \frac{1}{ \sqrt{n}} ln(n) = \lim_{ n \to \infty } \frac{ \frac{1}{n} } { \frac{1}{2 \sqrt{n} } } = \lim_{ n \to \infty } \frac{2 \sqrt{n} }{n} = \lim_{ n \to \infty } \frac{2}{ \sqrt{n} } =0 $

    Remember that this was the limit of $\displaystyle ln(y) $. So what we can do is $\displaystyle e^{ln(y)} = y $

    Thus,

    $\displaystyle \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } = \lim_{ n \to \infty } y = \lim_{ n \to \infty } e^{ln(y)} $ => $\displaystyle e^ { \lim_{ n \to \infty } ln(y) } = e^0 = 1 $
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Apr 2010
    Posts
    30
    Ok, so can L'hopitals rule always be used to find the limit of sequences?

    Like if i had {(2n+3)(3n+2)}/{(n^2)+3n+2} would the limit be 6?

    Also if i had {log(n)}/{10th root of n}..... sorry latex fail again, would i be able to use L'hopitals rule this time?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by Mathman87 View Post
    Ok, so can L'hopitals rule always be used to find the limit of sequences?

    Like if i had {(2n+3)(3n+2)}/{(n^2)+3n+2} would the limit be 6?

    Also if i had {log(n)}/{10th root of n}..... sorry latex fail again, would i be able to use L'hopitals rule this time?
    You can use L'Hopitals anytime there is a limit of indeterminate form. And you cannot ask what the limit of a function or a series of functions will be without providing where our limit is taken (i.e. x--> infinity or x-->2).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2010
    Posts
    30
    Ok i see. So i tried using L'hopitals rule for {log(n)}/{n^(1/10)}

    So i differentiated the denominator and the numerator and ended up with 10*{n^(-1/10)}. Is this right?

    Im not convinced im using it right lol!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by Mathman87 View Post
    Ok i see. So i tried using L'hopitals rule for {log(n)}/{n^(1/10)}

    So i differentiated the denominator and the numerator and ended up with 10*{n^(-1/10)}. Is this right?

    Im not convinced im using it right lol!
    Dude, you need to add limits!

    $\displaystyle \frac {log(n)} {n^{1/10} } $

    This is an equation and L'Hopitals has nothing to do with it!

    However, if we have

    $\displaystyle \lim_{ n \to \infty } \frac {log(n)} {n^{1/10} }$

    This is of the form $\displaystyle \frac { \infty }{ \infty } $ and we can use L'hopitals.

    $\displaystyle \lim_{ n \to \infty } \frac {log(n)} {n^{1/10} } = \lim_{ n \to \infty } \frac{ \frac{1}{n} }{ \frac{1}{10 n^{ \frac{9}{10} } } } = \lim_{ n \to \infty } \frac{ 10 n^{ \frac{9}{10} } }{n} $

    $\displaystyle 10 \lim_{ n \to \infty } \frac{1}{n^{ \frac{1}{10} } } = 0$
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Apr 2010
    Posts
    30
    I know, but it doesnt tell us what n is tending towards on the question, hence why im a bit confused?
    Iv attatched the document so you can see what i mean. Its question 4....
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    Since this is a test from 2007/2008, I would assume the prof. may have stated what the limits are approaching in class.

    4
    (i) is probably infinity and then the answer is e
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    (iii) proabably 0
    $\displaystyle \frac{6n^2+11n+6}{n^2+3n+2}\rightarrow \lim_{n\to 0}\frac{6n^2}{n^2}=6$

    (iv)+(v)
    AllanCuz answered

    (ii) is probably supposed to be approaching infinity
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Sequence Limits
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Aug 13th 2010, 07:22 AM
  2. [SOLVED] Problem on limits of a sequence.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 20th 2010, 06:25 PM
  3. Replies: 2
    Last Post: Mar 1st 2010, 11:57 AM
  4. Sequence Limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 23rd 2009, 06:32 PM
  5. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Apr 26th 2009, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum