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Math Help - limits of a sequence

  1. #1
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    limits of a sequence

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

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

    Sorry about latex fail, its n to the power the whole of the bracket.
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    Quote Originally Posted by Mathman87 View Post
    In general, how do you go about finding the limit of a sequence?

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

    Sorry about latex fail, its n to the power the whole of the bracket.
    What is n approaching?
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  3. #3
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    It doesnt specify on the examples, but i imagine infinity?
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  4. #4
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    \infty^0 is an indeterminant form. Do you know how to solve those?
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  5. #5
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    no?
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  6. #6
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    L'Hopital's Rule
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    Quote Originally Posted by Mathman87 View Post
    In general, how do you go about finding the limit of a sequence?

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

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

    Let,

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

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

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

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

      \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  ln(y) . So what we can do is  e^{ln(y)} = y

    Thus,

     \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } = \lim_{ n \to \infty } y = \lim_{ n \to \infty } e^{ln(y)} =>  e^ { \lim_{ n \to \infty } ln(y) } = e^0 = 1
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    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?
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  9. #9
    Senior Member AllanCuz's Avatar
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    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).
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  10. #10
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    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!
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  11. #11
    Senior Member AllanCuz's Avatar
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    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!

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

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

    However, if we have

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

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

     \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}

     10 \lim_{ n \to \infty } \frac{1}{n^{ \frac{1}{10} } } = 0
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  12. #12
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    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
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  13. #13
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    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
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  14. #14
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    (iii) proabably 0
    \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
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