Evaluation of $ \displaystyle \lim_{x \to 0} \frac{\sin ( \pi \cos(x) )}{x \sin x}$ using a Taylor Approximation 
Compute the limit: $$ \displaystyle \lim_{x \to 0} \frac{\sin ( \pi \cos(x) )}{x \sin (x)}$$
  using a Taylor approximation 

It seems very intuitive for me to use a Taylor Approximation for this kind of a problem, since the values of $x$ get arbitrary small. 
My Approach (wrong): 
$$\cos (x)= 1 - \frac{x^2}{2}+O(x^4) \implies \pi \cos (x)= \pi - \frac{\pi x^2}{2}+ \dots  \\ \sin (x) = x - \frac{x^3}{6}+O(x^6)$$
My problem now is to compute $\sin ( \pi \cos (x)) $. One of my tutors once told me for small values of $x$ I can always use $\sin (x) \sim x$, so I thought that the following must be true as well for small values of $x$: $$\sin (\pi \cos (x))= \sin\left(\pi - \frac{\pi x^2}{2}+ \dots \right) \sim \pi - \frac{\pi x^2}{2}+ \dots$$
Which is indeed very wrong, the correct answer is:
$$\sin (\pi \cos (x))\sim \frac{\pi x^2}{2}\pm O(x^4)$$
While it seems clear to me that my answer cannot be true, (I did it by naively plugging in the values and obtained the same result as I already had for $\pi \cos (x)$), I don't see how the above result is obtained. Could somebody show me how to correctly process in that manner?
Note: In class we were not yet introduced to the $O$-Notation. In the book by C.T. Michaels he simply uses dots for the higher terms and says that they can be neglected. I tried to include the $O$-Notation in my question, but only with my intuitive understanding of it. 
 A: You've done everything almost right; the problem is that 
$$\sin {(\pi - \frac{\pi x^2}{2})} \nsim \pi - \frac{\pi x^2}{2}$$ beacuse the argument of the sin does not goes to zero! (it goes to $\pi$!)
The correct way of doing that is 
$$ \sin {(\pi - \frac{\pi x^2}{2})} = \sin {\frac{\pi x^2}{2}} \sim \frac{\pi x^2}{2}$$
(becasue $\sin(\pi - x) = \sin x$); note that this time the argument of the sin actually goes to zero
A: HINT:
We can avoid Taylor's expansion as follows
As $\sin(\pi-y)=\sin y$
$\sin(\pi\cos x)=\sin\{\pi-\pi\cos x\}=\sin\{\pi(1-\cos x)\}$
A: Your approach failed, because you aren't plugging small values into $\sin x$: you're plugging in values near $\pi$.
In order to use that approach, you need to use an identity to transform the problem into one where you are plugging something small into $\sin$.
Incidentally, formally, we have (as $x \to 0$):
$$ \sin x = x + O(x^3) $$
And also
$$ (\pi - \frac{\pi}{2} x^2  + O(x^4))^3
 = (\pi + O(x))^3
 = \pi^3 + O(x)$$
so the thing you plugged in gives
$$ \begin{align}\sin(\pi \cos x) &= \sin(\pi - \frac{\pi}{2} x^2  + O(x^4))
\\ &= \pi - \frac{\pi}{2} x^2 + O(x^4) + O(\pi^3 + O(x)) = O(\pi^3)
\end{align}$$
which is a particularly useless approximation.
A: Hint:
$sin( \pi cos(x) )\approx  \frac {\pi }{2}x^2  -\frac {\pi}{24} x^4 +\frac {\pi -15\pi^3} {720}x^6 +O(x^8) $
