A limit question related to the nth derivative of a function 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.
 A: 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 $f(x) = 1-\e + O(\e^2) = 1-\frac{x^2}{6} + O(x^4)$. 
A: 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
$$
\begin{align}
f(0)&=1\\
f''(0)&=-\frac13\\
f^{(4)}(0)&=-\frac{7}{15}\\
&\text{etc.}
\end{align}
$$
