Results 1 to 10 of 10

Thread: Sequences - Limits

  1. #1
    Junior Member pearlyc's Avatar
    Joined
    Apr 2008
    Posts
    70

    Sequences - Limits

    Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

    Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence $\displaystyle {a_n}$ if it exists. Otherwise explain why the sequence diverges.

    (a) $\displaystyle a_n$ = $\displaystyle \frac{log n + 5n^2}{2n^2 + 100}$

    I divided the whole thing with $\displaystyle n^2$ and using standard limits I got,

    $\displaystyle \frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

    Then I used l'Hopital's rule to solve the limit for the $\displaystyle \frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.

    (b) $\displaystyle a_n = \frac{3^n + n!}{100^n + n^7}$

    I don't know how to do this one!

    (c) $\displaystyle a_n = (2^n + 1)^\frac{1}{n}$

    This one is to use sandwich rule right?

    (d) $\displaystyle a_n = cos(\frac {\pi n}{3n + 5})$

    Used continuity rule and got,

    $\displaystyle cos(\frac {lim \pi n}{lim 3n + 5})$
    $\displaystyle cos(\frac { \pi lim n}{3 lim n + lim 5})$

    divided the whole thing by n

    $\displaystyle cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
    $\displaystyle cos(\frac {\pi}{3})$
    = 0.5

    (e) $\displaystyle a_n = n tan(\frac{1}{n})$

    I don't know how to do this one either.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello

    Quote Originally Posted by pearlyc View Post
    Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

    Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence $\displaystyle {a_n}$ if it exists. Otherwise explain why the sequence diverges.

    (a) $\displaystyle a_n$ = $\displaystyle \frac{log n + 5n^2}{2n^2 + 100}$

    I divided the whole thing with $\displaystyle n^2$ and using standard limits I got,

    $\displaystyle \frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

    Then I used l'Hopital's rule to solve the limit for the $\displaystyle \frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.
    L'H˘pital's rule would yield :

    $\displaystyle \frac{\frac 1n}{2n}=\frac{1}{2n^2}$, and the limit of this is 0.



    don't havd time for thinking about the following ones, sorry
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by pearlyc View Post
    (d) $\displaystyle a_n = cos(\frac {\pi n}{3n + 5})$

    Used continuity rule and got,

    $\displaystyle cos(\frac {lim \pi n}{lim 3n + 5})$
    $\displaystyle cos(\frac { \pi lim n}{3 lim n + lim 5})$

    divided the whole thing by n

    $\displaystyle cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
    $\displaystyle cos(\frac {\pi}{3})$
    = 0.5
    This is right

    I will answer the remaining in a while..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by pearlyc View Post
    Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

    Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence $\displaystyle {a_n}$ if it exists. Otherwise explain why the sequence diverges.

    (a) $\displaystyle a_n$ = $\displaystyle \frac{log n + 5n^2}{2n^2 + 100}$

    I divided the whole thing with $\displaystyle n^2$ and using standard limits I got,

    $\displaystyle \frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

    Then I used l'Hopital's rule to solve the limit for the $\displaystyle \frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.

    (b) $\displaystyle a_n = \frac{3^n + n!}{100^n + n^7}$

    I don't know how to do this one!

    (c) $\displaystyle a_n = (2^n + 1)^\frac{1}{n}$

    This one is to use sandwich rule right?

    (d) $\displaystyle a_n = cos(\frac {\pi n}{3n + 5})$

    Used continuity rule and got,

    $\displaystyle cos(\frac {lim \pi n}{lim 3n + 5})$
    $\displaystyle cos(\frac { \pi lim n}{3 lim n + lim 5})$

    divided the whole thing by n

    $\displaystyle cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
    $\displaystyle cos(\frac {\pi}{3})$
    = 0.5

    (e) $\displaystyle a_n = n tan(\frac{1}{n})$

    I don't know how to do this one either.
    For b)

    Remember that the factorial function grows faster than polynomials or exponentials. The series diverges .


    For c)

    try this trick

    $\displaystyle a_n = (2^n + 1)^\frac{1}{n}$ Take the natural log of both sides

    $\displaystyle \ln(a_n) = \ln \left((2^n + 1)^\frac{1}{n}\right))$

    using log properties we get

    $\displaystyle \ln(a_n)=\frac{1}{n} \cdot \ln(2^n+1) =\frac{\ln(2^n+1)}{n}$
    We can now use L'hospitals rule to get

    $\displaystyle \ln(a_n)=\frac{\frac{(\ln(2))2^n}{2^n+1}}{1}$

    Now letting n go to infinity gives

    $\displaystyle \ln(a_n)=\ln(2) \iff a_n=2$

    For d you are correct

    for e) rewrite as $\displaystyle \frac{\tan(\frac{1}{n})}{\frac{1}{n}}$

    and use L.H rule

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi
    (c) $\displaystyle a_n = (2^n + 1)^\frac{1}{n}$

    This one is to use sandwich rule right?
    Yes, you can use the squeeze theorem : $\displaystyle 2^{n+1} > 2^n+1 > 2^n$ hence ...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member pearlyc's Avatar
    Joined
    Apr 2008
    Posts
    70
    Whoa, thanks for the many responses guys (: Took your guidance and attempted the questions!

    For (c), this is how far I've got ..

    $\displaystyle \sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}$
    $\displaystyle lim (2)^\frac{1}{n} < \sqrt[n]{2^n + 1} < lim 2.2^\frac{1}{n}$
    $\displaystyle 1 < \sqrt [n]{2^n+1} < 2$

    Where do I go from here? Hmm.

    As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

    After differentiating, $\displaystyle \frac {-1}{x^2} sec^2x$

    I don't know where to go from here too
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by pearlyc View Post
    $\displaystyle \sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}$
    $\displaystyle lim (2)^\frac{1}{n} < \sqrt[n]{2^n + 1} < lim 2.2^\frac{1}{n}$

    You missed a minor point.
    Try again... what is $\displaystyle \sqrt[n]{2^n}$


    As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

    After differentiating, $\displaystyle \frac {-1}{x^2} sec^2x$

    I don't know where to go from here too
    LH rule can be a little dangerous here. Instead modify the question a little bit and see if you can recognize the limit...

    When $\displaystyle n \to \infty, \frac1{n} \to 0$, so lets call $\displaystyle \frac1{n}$ as $\displaystyle \theta$.

    Now where have I seen $\displaystyle \lim_{\theta \to 0} \frac{\tan \theta}{\theta}$?

    If you havent seen this limit before, LH is still an option on this...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member pearlyc's Avatar
    Joined
    Apr 2008
    Posts
    70
    Oh OOPS!

    Thanks, hahaha.

    Eh, I still don't really get that tan question!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by pearlyc View Post
    Oh OOPS!

    Thanks, hahaha.

    Eh, I still don't really get that tan question!
    Limit(n to infinity) is equivalent to

    Try L'Hospitals rule
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by pearlyc View Post
    As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

    After differentiating, $\displaystyle \frac {-1}{x^2} sec^2x$

    I don't know where to go from here too
    Well part of the problem is you shouldn't have ended up here.

    Note that the orginial sequence was

    $\displaystyle n\tan\left( \frac{1}{n}\right)$

    Rewriting as $\displaystyle \frac{\tan\left( \frac{1}{n}\right)}{\frac{1}{n}}$

    as $\displaystyle n \to \infty$ this goes to $\displaystyle \frac{0}{0}$

    Now applying L'hospitials rule we get

    $\displaystyle \frac{\sec^{2}(\frac{1}{n})\cdot (\frac{-1}{n^2})}{(\frac{-1}{n^2})}$

    This is where your error occured you forgot to use the chain rule when taking the derivative of the tangent function

    Now when we reduce we get

    $\displaystyle \lim_{n \to \infty}{\sec^{2}\left( \frac{1}{n}\right)} \to \sec^2(0)=1^2=1 $

    I hope this clears it up.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limits of sequences
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Oct 11th 2011, 10:33 AM
  2. Sequences and limits
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 1st 2011, 07:00 PM
  3. Sequences - limits
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Oct 1st 2008, 03:17 AM
  4. Limits and sequences
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 3rd 2008, 03:51 PM
  5. limits of sequences
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 4th 2007, 05:38 PM

Search Tags


/mathhelpforum @mathhelpforum