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Math Help - need help finding a limit

  1. #1
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    need help finding a limit

    Hi, this limit is confusing me can someone please explain this one to me:
    Limit as k goes to infinity k^2*sin^2(1/k).
    Thanks for the help
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  2. #2
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    kČ = infinity * infinity = infinity
    1/k = 0, because divide through infinity gives 0

    sin (0) = 0

    sinČ (o) = 0Č = 0

    0 * kČ = 0 * infinity = NOT DEFINED
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by cowboys111 View Post
    Hi, this limit is confusing me can someone please explain this one to me:
    Limit as k goes to infinity k^2*sin^2(1/k).
    Thanks for the help
    Substitute : u=1/k.

    When k goes to infinity, u goes to 0.

    \lim_{k \to \infty} k^2 \sin^2(1/k)=\lim_{u \to 0} \frac{\sin^2 u}{u^2}

    =\lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2=\lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2

    But \lim_{u \to 0} \frac{\sin u-\sin 0}{u-0} is the value of the derivative of \sin u when u equals 0.
    The derivative of \sin u is \cos u


    Therefore, \lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2=(\cos 0)^2=1

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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cowboys111 View Post
    Hi, this limit is confusing me can someone please explain this one to me:
    Limit as k goes to infinity k^2*sin^2(1/k).
    Thanks for the help
    Alternately

    L=\lim_{n\to\infty}n^2\sin^2\bigg(\frac{1}{n}\bigg  )=\lim_{n\to\infty}\bigg[n\sin\bigg(\frac{1}{n}\bigg)\bigg]^2

    So then

    \sqrt{L}=\lim_{n\to\infty}n\sin\bigg(\frac{1}{n}\b  igg)=\lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\  bigg)}{\frac{1}{n}}

    Note this is an indetermiate form so we can apply L'hopitals

    so \lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\bigg)  }{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{-1}{n^2}\cos\bigg(\frac{1}{n}\bigg)}{\frac{-1}{n^2}}=\lim_{n\to\infty}\cos\bigg(\frac{1}{n}\bi  gg)=1

    So \sqrt{L}=1\Rightarrow{L=1^2=1}
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  5. #5
    Moo
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    Note this is an indetermiate form so we can apply L'hopitals
    I was waiting for this, I knew you would reply by using it !
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    I was waiting for this, I knew you would reply by using it !
    ...well I was going to do it your way but mid-type I saw you did it! So blame yourself
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  7. #7
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    But having n\sin \frac{1}{n} as n\to\infty we only need a substitution and we'll get a well known limit.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    But having n\sin \frac{1}{n} as n\to\infty we only need a substitution and we'll get a well known limit.
    If your statement is directed at me, I usually try to do the weird abstract method that is easier but harder to see. But when I am beaten to it I show the more commonly known method such as L'hopital's. Substitution in limits can be a foreign concept to most high schoolers/college students so I showed this method to highlight how the limit is probably meant to be solved in their class or hwo to apply a known theorem.
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  9. #9
    Math Engineering Student
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    It's not, actually, directed to ya, I said it as another way to tackle the problem.

    And as for the "foreign" concept, that's false, it depends how one does teach such technique.
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