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.

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 $f(x) = 1-\e + O(\e^2) = 1-\frac{x^2}{6} + O(x^4)$. 
