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Thread: hint for limit question

  1. #1
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    hint for limit question

    I'm puzzled with the following limit: $\displaystyle \lim_{n \rightarrow \infty } {\sqrt[n]{{b}^{ \frac{1}{ 2^{n} } } - 1}} $ and $\displaystyle b > 1$
    the exponent is $\displaystyle \frac{1}{2^n}$
    It is easy to see that the limit lies in [0,1] interval.
    Using online graph plotter I can see that the answer is $\displaystyle \frac{1}{2}$ but no matter what I try I can't reach this answer analytically.
    I tried quite a few manipulations like: extrating $\displaystyle \frac{1}{2^{n}}$ factor, geometric sum, representing the formula recursively through $\displaystyle a_n$ and $\displaystyle a_{n+1}$ and some other symbol manipulations.

    Can someone point me to the promising direction ?
    Thank you
    Last edited by ktotam; Oct 31st 2013 at 07:07 AM.
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  2. #2
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    Re: hint for limit question

    Quote Originally Posted by ktotam View Post
    I'm puzzled with the following limit: $\displaystyle \lim_{n \rightarrow \infty } {\sqrt[n]{{b}^{ \frac{1}{ 2^{n} } } - 1}} $ and $\displaystyle b > 1$
    the exponent is $\displaystyle \frac{1}{2^n}$
    It is easy to see that the limit lies in [0,1] interval.
    Using online graph plotter I can see that the answer is $\displaystyle \frac{1}{2}$ but no matter what I try I can't reach this answer analytically.
    I tried quite a few manipulations like: extrating $\displaystyle \frac{1}{2^{n}}$ factor, geometric sum, representing the formula recursively through $\displaystyle a_n$ and $\displaystyle a_{n+1}$ and some other symbol manipulations.

    Can someone point me to the promising direction ?
    Thank you
    $\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty} \sqrt[n]{b^{\frac{1}{2n}} - 1} &= \lim_{n \to \infty} \left[ \left( b^{\frac{1}{2n}} - 1 \right) ^{\frac{1}{n}} \right] \\ &= \lim_{n \to \infty} e^{\ln{ \left[ \left( b^{\frac{1}{2n}} - 1 \right) ^{\frac{1}{n}} \right] }} \\ &= \lim_{n \to \infty} e^{ \frac{\ln{ \left( b^{\frac{1}{2n}} - 1 \right) }}{n} } \\ &= e^{\lim_{n \to \infty} \frac{\ln{ \left( b^{\frac{1}{2n}} - 1 \right) }}{n}} \end{align*}$

    and this is now in a form where you can apply L'Hospital's Rule.
    Thanks from topsquark and ktotam
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  3. #3
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    Re: hint for limit question

    Hi, Prove It

    Notice that the exponent of b is $\displaystyle \frac{1}{2^n}$ rather than $\displaystyle \frac{1}{2 \cdot n}$.

    Anyway I tried what you suggested with $\displaystyle \frac{1}{2^n}$ but it leads to: $\displaystyle \frac{ln{2} \cdot ln{b} \cdot b^{\frac{1}{2^n}}}{2^n \cdot (b^{\frac{1}{2^n}}-1)}$.
    It is more complex expression for n so probably not a way to solve it.
    By the way n is discrete variable.
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  4. #4
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    Re: hint for limit question

    Quote Originally Posted by ktotam View Post
    Hi, Prove It

    Notice that the exponent of b is $\displaystyle \frac{1}{2^n}$ rather than $\displaystyle \frac{1}{2 \cdot n}$.

    Anyway I tried what you suggested with $\displaystyle \frac{1}{2^n}$ but it leads to: $\displaystyle \frac{ln{2} \cdot ln{b} \cdot b^{\frac{1}{2^n}}}{2^n \cdot (b^{\frac{1}{2^n}}-1)}$.
    It is more complex expression for n so probably not a way to solve it.
    By the way n is discrete variable.
    Prove It is correct. Also, you missed a negative sign. You have:

    $\displaystyle \begin{align*}\lim_{n\to \infty} \sqrt[n]{b^{2^{-n}}-1} & = \exp\left(\lim_{n\to \infty} \dfrac{\ln\left(b^{2^{-n}}-1\right)}{n}\right) \\ & = \exp\left(\lim_{n\to \infty}-\dfrac{\ln 2\cdot \ln b \cdot b^{2^{-n}}}{2^n\cdot (b^{2^{-n}}-1)}\right) \\ & = \exp\left(\lim_{n\to \infty}-\dfrac{\ln 2\cdot \ln b \cdot 2^{-n}}{1-b^{-2^{-n}}}\right)\end{align*}$

    You now have another form where you can apply L'Hospital's Rule again. It doesn't matter if n is discrete or not. L'Hospital's still applies.
    Last edited by SlipEternal; Oct 31st 2013 at 09:37 AM.
    Thanks from ktotam
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  5. #5
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    Re: hint for limit question

    I verified that after applying L'Hospital's rule second time I get the desired limit.
    Thanks you
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