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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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    @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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    @EugeneShva$r$ts: 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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    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