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Math Help - Sequences - Limits

  1. #1
    Junior Member pearlyc's Avatar
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    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 {a_n} if it exists. Otherwise explain why the sequence diverges.

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

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

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

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

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

    I don't know how to do this one!

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

    This one is to use sandwich rule right?

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

    Used continuity rule and got,

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

    divided the whole thing by n

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

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

    I don't know how to do this one either.
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  2. #2
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    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 {a_n} if it exists. Otherwise explain why the sequence diverges.

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

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

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

    Then I used l'Hopital's rule to solve the limit for the \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 :

    \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
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    (d) a_n = cos(\frac {\pi n}{3n + 5})

    Used continuity rule and got,

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

    divided the whole thing by n

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

    I will answer the remaining in a while..
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  4. #4
    Behold, the power of SARDINES!
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    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 {a_n} if it exists. Otherwise explain why the sequence diverges.

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

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

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

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

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

    I don't know how to do this one!

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

    This one is to use sandwich rule right?

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

    Used continuity rule and got,

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

    divided the whole thing by n

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

    (e) 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

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

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

    using log properties we get

    \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

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

    Now letting n go to infinity gives

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

    For d you are correct

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

    and use L.H rule

    Good luck.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Hi
    (c) a_n = (2^n + 1)^\frac{1}{n}

    This one is to use sandwich rule right?
    Yes, you can use the squeeze theorem : 2^{n+1} > 2^n+1 > 2^n hence ...
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  6. #6
    Junior Member pearlyc's Avatar
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    Whoa, thanks for the many responses guys (: Took your guidance and attempted the questions!

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

    \sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}
    lim (2)^\frac{1}{n} < \sqrt[n]{2^n + 1} < lim 2.2^\frac{1}{n}
    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, \frac {-1}{x^2} sec^2x

    I don't know where to go from here too
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by pearlyc View Post
    \sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}
    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 \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, \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 n \to \infty, \frac1{n} \to 0, so lets call \frac1{n} as \theta.

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

    If you havent seen this limit before, LH is still an option on this...
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  8. #8
    Junior Member pearlyc's Avatar
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    Oh OOPS!

    Thanks, hahaha.

    Eh, I still don't really get that tan question!
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  9. #9
    Lord of certain Rings
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    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
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  10. #10
    Behold, the power of SARDINES!
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    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, \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

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

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

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

    Now applying L'hospitials rule we get

    \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

    \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.
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