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Calculate the limit $ \displaystyle \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $

I now the answer, it is $ \displaystyle e^\frac{\log^2z}{2} $, but I don't know how to prove it. It seems like this notable limit $\displaystyle \lim \limits_{x \to \infty} {\left(1 + \frac{c}{x}\right)^x} = e^c$ should be useful here. For example I tried this way: $$ (z^{1/\sqrt n} + z^{-1/\sqrt n}) = (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 + 2 $$

$$ \displaystyle \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} = \displaystyle \lim \limits_{n \to \infty} {\left(1 + \frac{(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2}{2}\right)^n} $$

where $ (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 $ seems close to $ \frac{\log^2 z}{n} $.

Also we can say that $$ \left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n = e^{n \log {\left(1 + \frac{\left(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)}\right)^2}{2}\right)}}$$ and $ \log {\left(1 + \frac{(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2}{2}\right)} $ can be expand in the Taylor series. But I can't finish this ways.

Thanks for the help!

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    $\begingroup$ If you are going to use Taylor series, it is useful to let $z=e^t$. The same is true if you are going to use L'Hospital's Rule. $\endgroup$ – André Nicolas Feb 27 '16 at 16:04
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Assume $z>0$. One may write, as $n \to \infty$, $$ \begin{align} z^{1/\sqrt n}=e^{(\log z)/\sqrt n}&=1+\frac{\log z}{\sqrt n}+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right)\\ z^{-1/\sqrt n}=e^{-(\log z)/\sqrt n}&=1-\frac{\log z}{\sqrt n}+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right) \end{align} $$ giving $$ \frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}=1+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right) $$ and, as $n \to \infty$, $$ \begin{align} \left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n&=\left(1+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right)\right)^n\\\\ &=e^{(\log z)^2/2}+O\left(\frac1{n^{1/2}} \right) \to e^{(\log z)^2/2} \end{align} $$

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Set $a = \log z$. Using the continuity of $\log$ we take the logarithm of the limit, so that we only need to calculate \begin{equation} \lim_{n\to \infty} n\log(\frac{e^{a\frac{1}{\sqrt{n}}}+e^{-a\frac{1}{\sqrt{n}}}}{2}) = \lim_{n\to \infty} n \log \cosh(\frac{a}{\sqrt{n}}). \end{equation} We have \begin{equation} \lim_{x\to 0}\frac{\log\cosh(ax)}{x^2} = \lim_{x\to 0}\frac{\frac{a\sinh(ax)}{\cosh(ax)}}{2x} = \frac{a}{2}\lim_{x\to 0}\frac{\sinh(ax)}{x} = \frac{a^2}{2}, \end{equation} Hence \begin{equation} \lim_{n\to \infty} n \log \cosh(\frac{a}{\sqrt{n}})= a^2/2. \end{equation} Taking the exponent we obtain the result.

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$e^x$ = $1 + x + x^2 / 2 + x^3 / 3! + x^4 / 4! ...$

$e^{-x}$ = 1 - x + x^2 / 2 - x^3 / 3! + x^4 / 4! - + ...$

$((e^x + e^{-x}) / 2)$ = $1 + x^2 / 2 + x^4 / 4! + x^6 / 6! ...

In your case, x = $ln z / n^{1/2}$, which makes

$((e^x + e^{-x}) / 2)$ = $1 + ln^2 z / 2n + ln^4 z / 4!n^2 + ln^6 z / 6!n^3 ... $

To raise to the n-th power, we take the logarithm, about $ln^2 z / 2n + O (1/n^2)$, multiply by n giving about $ln^2 z / 2 + O (1/n)$, exponentiate giving about $exp(ln^2 z / 2 + O (1/n))$ so the limit is $exp(ln^2 z / 2)$ or $z^ {ln z / 2}$

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If $L$ is the desired limit then we have \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\right\}\notag\\ &= \lim_{n \to \infty}\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\ &= \lim_{n \to \infty}n\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\dfrac{\log\left(1 + \dfrac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}} - 2}{2}\right)}{\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}}\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\ &= \lim_{n \to \infty}n\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\ &=\frac{1}{2}\lim_{n \to \infty}n\left(\frac{z^{1/\sqrt{n}} - 1}{z^{1/2\sqrt{n}}}\right)^{2}\notag\\ &= \frac{1}{2}\lim_{n \to \infty}\{\sqrt{n}(z^{1/\sqrt{n}} - 1)\}^{2}\notag\\ &= \frac{(\log z)^{2}}{2}\notag \end{align} Hence $L = \exp\left\{\dfrac{(\log z)^{2}}{2}\right\}$.

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  • $\begingroup$ Very well done! +1 $\endgroup$ – Mark Viola Feb 28 '16 at 4:18
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HINT:

If one wants to uses asymptotic analysis, then simply note that

$$\frac{z^{1/\sqrt n}+z^{-1/\sqrt n}}{2}=\cosh\left(\frac{\log(z)}{\sqrt n}\right)=1+\frac12 \frac{\log^2(z)}{n}+O(1/n^2)$$


Alternatively, we can proceed using only elementary inequalities, established without calculus, and the squeeze theorem. To proceed, we write

$$\begin{align} \cosh^n\left(\frac{\log(z)}{\sqrt n}\right)&=e^{n\log\left(\cosh\left(\frac{\log(z)}{\sqrt n}\right)\right)}\\\\&=e^{n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)} \tag 1\\\\ \end{align}$$

Now, in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that for $|x|<1$

$$1+x\le e^{x}\le \frac{1}{1-x} \tag 2$$

and

$$\frac{x}{1+x}\le\log(1+x)\le x \tag 3$$

From $(1)$ we can easily show that

$$\frac{x}{1-x^2}\le \sinh(x)\le x \tag 4$$

for $0\le x <1$.

Applying $(3)$ and $(4)$ to $(1)$ reveals

$$\frac{\frac12 \log^2(z)}{\left(1-\left(\frac{\log(z)}{2\sqrt n}\right)^2\right)^2} \left(\frac{1}{\cosh\left(\frac{\log(z)}{\sqrt n}\right)}\right)\le n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)\le \frac12 \log^2(z)$$

whereupon applying the squeeze theorem reveals

$$\lim_{n\to \infty}n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)=\frac12 \log^2(z)$$

Finally, exploiting the continuity of the exponential function yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{z^{1/\sqrt n}+z^{-1/\sqrt n}}{2}=e^{\frac12 \log^2(z)}}$$

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