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This evening I thought of the following question that isn't related to homework, but it's a question that seems very challenging to me, and I take some interest in it.

Let's consider the following function: $$ f(x)= \left(\frac{\sin x}{x}\right)^\frac{x}{\sin x}$$ I wonder what is the first derivative (1st, 2nd, 3rd ...) such that $\lim\limits_{x\to0} f^{(n)}(x)$ is different from $0$ or $+\infty$, $-\infty$, where $f^{(n)}(x)$ is the nth derivative of $f(x)$ (if such a case is possible). I tried to use W|A, but it simply fails to work out such limits. Maybe i need the W|A Pro version.

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    Gotta bite. What is 'W|A'?2012-07-08
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    @copper.hat I assume Wolfram Alpha2012-07-08
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    @copper.hat: hi. Wolfram|Alpha2012-07-08
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    Thanks. New form of abbreviation for me.2012-07-08
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    I just want to point out that this function fails to be defined in many cases. For example, for any natural number $k\in\mathbb{N}$, let $x=\frac{(3+4k)\pi}{2}$. Then we have $$f(x)=\left(\frac{\sin(x)}{x}\right)^\frac{x}{\sin(x)}= \left(\frac{-1}{x}\right)^{\frac{x}{-1}}=\left(-\frac{1}{x}\right)^{-x}$$ which is a negative number raised to a negative power. Any positive $x$ for which $\sin(x)$ is negative, or vice versa, will have the same problem. Similarly, for any integer $k\in\mathbb{Z}$, let $x=k\pi$. Then we are raising $0$ to the $\frac{x}{0}$ power, which I would say is undefined.2012-07-08
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    @Zev Chonoles: i've found a formula to make W|A give me the answer for 2nd derivative. It seems to be $-\frac{1}{3}$. I'd like so much to know if there is a simple way to get this answer.2012-07-08
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    http://www.wolframalpha.com/input/?i=lim+x-%3E0+[%28sin%28x%29%2Fx%29^%28x%2Fsin%28x%29%29]%27%272012-07-08
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    @ZevChonoles If we fill in removable singularities, won't the function be well-defined (and differentiable) in a neighborhood of 0?2012-07-08
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    @Eugene: Ah, I'd misread the problem as to determine $\lim\limits_{x\to\infty}f^{(n)}(x)$, which my comment was pointing out wouldn't make sense. I think that $\lim\limits_{x\to0}f^{(n)}(x)$ should be okay.2012-07-08
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    @ZevChonoles looks like it works out in the end :]2012-07-08
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    @Chris'sister: You can try [this W|A link](http://www.wolframalpha.com/input/?i=Series[%28Sin[x]%2Fx%29^%28x%2FSin[x]%29%2C+{x%2C+0%2C+10}]) to see the power series and then multiply the coefficient of $x^n$ by $n!$ to get $f^{(n)}(0)$.2012-07-08

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The Taylor expansion is $$f(x) = 1 - \frac{x^2}{6} + O(x^4),$$ so \begin{eqnarray*} f(0) &=& 1 \\ f'(0) &=& 0 \\ f''(0) &=& -\frac{1}{3}. \end{eqnarray*}

$\def\e{\epsilon}$

Addendum: We use big O notation. Let $$\e = \frac{x}{\sin x} - 1 = \frac{x^2}{6} + O(x^4).$$ Then \begin{eqnarray*} \frac{1}{f(x)} &=& (1+\e)^{1+\e} \\ &=& (1+\e)(1+\e)^\e \\ &=& (1+\e)(1+O(\e\log(1+\e))) \\ &=& (1+\e)(1+O(\e^2)) \\ &=& 1+\e + O(\e^2), \end{eqnarray*} so $f(x) = 1-\e + O(\e^2) = 1-\frac{x^2}{6} + O(x^4)$.

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    Isn't that just the Taylor series for $\dfrac{\sin x}{x}$? I understand the exponent goes to 1 as $x \to 0$, but is that approach valid from the naive perspective?2012-07-08
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    @EugeneShvarts: The series happen to agree to order $x^2$. There is a difference at order $x^4$.2012-07-08
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    @Chris'sister: Glad to help. Cheers!2012-07-08
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    @oen Ah, okay. Did you realize this intuitively, or is there a quick technique to see this?2012-07-08
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    @oen: hold on! How did you get the Taylor expansion form??2012-07-08
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    @Chris'sister: I'll add something about that.2012-07-08
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    @oen: I was about to miss the most important part. :-) OK!2012-07-08
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    @EugeneShvarts: I added something about this to the answer.2012-07-08
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    @oen Thanks a lot -- that's very instructive.2012-07-08
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    @EugeneShvarts: You're welcome. Glad to help.2012-07-08
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First of all, note that $$ f(x)=\left(\frac{\sin(x)}{x}\right)^{\Large\frac{x}{\sin(x)}}\tag{1} $$ is an even function. This means that all the odd terms in the power series will be zero.

Using the power series for $\log(1+x)$, we get $$ \begin{align} &\log\left(\left(1-\frac16x^2+\frac{1}{120}x^4+O\left(x^6\right)\right)^{\Large1+\frac16x^2+\frac{7}{360}x^4+O\left(x^6\right)}\right)\\ &=\left(-\frac16x^2-\frac{1}{180}x^4+O\left(x^6\right)\right)\left(1+\frac16x^2+\frac{7}{360}x^4+O\left(x^6\right)\right)\\ &=-\frac16x^2-\frac{1}{30}x^4+O\left(x^6\right)\tag{2} \end{align} $$ Then we apply the power series for $e^x$ to get $$ f(x)=1-\frac16x^2-\frac{7}{360}x^4+O\left(x^6\right)\tag{3} $$ Of course, using more terms in the power series for $\dfrac{\sin(x)}{x}$ and $\dfrac{x}{\sin(x)}$, we could get more terms for $f(x)$.

To get the derivatives at $x=0$, you can just use the fact that the Taylor series near $0$ is $$ f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\tag{4} $$ to get that $f^{(n)}(0)=0$ for all odd $n$, and $$ \begin{align} f(0)&=1\\ f''(0)&=-\frac13\\ f^{(4)}(0)&=-\frac{7}{15}\\ &\text{etc.} \end{align} $$

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    thanks for your solution! You mean Maclaurin series, right?2012-07-08
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    A Maclaurin series is a Taylor series centered at $0$. So you *can* call this a Maclaurin series, but it is still a Taylor series (which is more general).2012-07-08