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Thread: Trigonometric limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Trigonometric limit?

    Evaluate:

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Evaluate:

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
    It would seem that:

    $\displaystyle \lim_{x\rightarrow 0+} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (+\infty)^{0} = 1$

    and

    $\displaystyle \lim_{x\rightarrow 0-} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (-\infty)^{0} = 1$

    so

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = 1$
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  3. #3
    Moo
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    Quote Originally Posted by sinewave85 View Post
    It would seem that:

    $\displaystyle \lim_{x\rightarrow 0+} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (+\infty)^{0} = 1$
    But this is not defined

    and

    $\displaystyle \lim_{x\rightarrow 0-} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (-\infty)^{0} = 1$
    How can something positive have a negative limit ?

    so

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = 1$
    ...
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by Moo View Post
    But this is not defined


    How can something positive have a negative limit ?


    ...
    Very sorry -- I forgot that $\displaystyle \infty^{0}$ was undefined. Thanks for catching that, Moo. I would hate to lead someone astray.
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  5. #5
    Super Member fardeen_gen's Avatar
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    How do we deal with $\displaystyle \infty^{0}$ forms? This problem has been giving me a headache.
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  6. #6
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    $\displaystyle \frac{\sin^2(x)}{\sin^2(x)} (\frac{1}{\sin^2(x)}+\frac{1}{2\sin^2(x)}+ \mbox(...) +\frac{1}{n \sin^2(x)})^{\sin^2(x)}=\frac{1}{\sin^2(x)}(H_n)^{ \sin^2(x)}=L$

    $\displaystyle ln (L) = ln(H_n)$

    Harmonic number - Wikipedia, the free encyclopedia
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  7. #7
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    Quote Originally Posted by fardeen_gen View Post
    Evaluate:

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
    Using the fact that $\displaystyle \sin x \approx x$, for small x, we can write:

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = $
    $\displaystyle \lim_{t \rightarrow \infty} \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)^{\frac1{t^2}}$

    Taking logs and apply LHospital's rule, we have to evaluate $\displaystyle \lim_{t \rightarrow \infty} \frac{\log \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)}{t^2}$ to get $\displaystyle \log n!$, Finally the answer is n!
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Using the fact that $\displaystyle \sin x \approx x$, for small x, we can write:

    $\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = $ $\displaystyle \lim_{t \rightarrow \infty} \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)^{\frac1{t^2}}$

    Taking logs and apply LHospital's rule, we have to evaluate $\displaystyle \lim_{t \rightarrow \infty} \frac{\log \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)}{t^2}$ to get $\displaystyle \log n!$, Finally the answer is n!
    After one application of L'Hospital's rule, I get

    $\displaystyle \lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}} $

    then it would appear you did the following

    $\displaystyle \lim_{t \to \infty} \frac {\big(\ln(1) +\ln(2) + \ln(3) + ... + \ln(n)\big)(1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}})}{ 1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}} $ ?

    $\displaystyle = \lim_{t \to \infty} \ln(1*2*3*...*n) $ which would be correct if the previous step were correct

    What am I missing?
    Last edited by Random Variable; Jun 11th 2009 at 11:00 AM.
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  9. #9
    Lord of certain Rings
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    Quote Originally Posted by Random Variable View Post
    After one application of L'Hospital's rule, I get

    $\displaystyle \lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}} $

    then it would appear you did the following

    $\displaystyle \lim_{t \to \infty} \frac {\big(\ln(1) +\ln(2) + \ln(3) + ... + \ln(n)\big)(1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}})}{ 1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}} $ ?

    $\displaystyle = \lim_{t \to \infty} \ln(1*2*3*...*n) $ which would be correct if the previous step were correct

    What am I missing?
    You are not missing anything...Its wrong and was hastily written(I applied $\displaystyle t \to 0$)

    Here is the solution:

    $\displaystyle \lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}} $

    $\displaystyle = \lim_{t \to \infty} \dfrac {0 + \ln(2)\left(\frac2{n}\right)^{t^{2}} + \ln(3)\left(\frac3{n}\right)^{t^{2}} + ... + \ln(n)}{\left(\frac1{n}\right)^{t^{2}} + \left(\frac2{n}\right)^{t^{2}} + \left(\frac3{n}\right)^{t^{2}} + ... + 1} $

    $\displaystyle =\ln(n) $
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