Difficult Limit Question $ \frac{0}{0}$ indeterminate form I am trying to solve this limit: 
$$\lim_{x\to 0}\frac{\pi\sin{x}-\sin{\pi x}}{x(\cos{x}-\cos(\pi x))}$$
I've tried solving it using l'hopital and conjugating.
Any solution, tip, or direction will be appreciated.
Thanks in advance,
Michael.
 A: By Taylor
$$\frac{\pi x-\dfrac{\pi x^3}{6}-\pi x-\dfrac{\pi^3x^3}6\cdots}{x\left(1-\dfrac{x^2}2-1+\dfrac{\pi^2x^2}2\cdots\right)}$$
A: For the denominator $$\cos x-\cos\pi x=\dfrac{x^2(\pi^2-1)}2\cdot\dfrac{\sin\dfrac{x(1+\pi)}2}{\dfrac{x(1+\pi)}2}\cdot\dfrac{\sin\dfrac{x(\pi-1)}2}{\dfrac{x(\pi-1)}2}$$
For the numerator 
$$\dfrac{\pi\sin x-\sin(\pi x)}{x^3}=\dfrac{\pi x-\sin\pi x}{x^3}-\pi\cdot\dfrac{x-\sin x}{x^3}$$
See Are all limits solvable without L'Hôpital Rule or Series Expansion
A: $\sin(x)= x - \frac{1}{6}x^3+o(x^3)$ so replacing $x$ with $\pi\,x$ you get 
$\sin(\pi\,x)= \pi\,x - \frac{\pi^3}{6}x^3+o(x^3)$
$\cos(x)= 1 - \frac{1}{2}x^2+o(x^2)$ so replacing $x$ with $\pi\,x$ you get $\cos(\pi x)= 1 - \frac{\pi^2}{2}x^2+o(x^2)$
The remaining part is just reporting in the formula and doing the calculacy. For the sinus, the terms in $x^1$ vanish and for the cosinus the terms in $x^0$ also vanish in the difference.
The only important thing is to have consistency in your $o(x^n)$ terms.
$$\frac{0+(\frac{-\pi}{6}+\frac{\pi^3}{6})\,x^3+o(x^3)}{x(0+(\frac{-1}{2}+\frac{\pi^2}{2})\,x^2+o(x^2))}=\frac{x^3(\frac{\pi(\pi^2-1)}{6}+o(1))}{x^3(\frac{(\pi^2-1)}{2}+o(1))}=\frac{\frac{\pi}{3}+o(1)}{1+o(1)}\to\frac{\pi}{3}$$

Otherwise, remember that L'Hospital rule is just a Taylor expansion in disguise.
If $f(0)=g(0)=0$, then $\frac{f(x)}{g(x)}=\frac{f(0)+xf'(0)+o(x)}{g(0)+xg'(0)+o(x)}=\frac{f'(0)+o(1)}{g'(0)+o(1)}\sim\frac{f'(0)}{g'(0)}$
If you still get a $0\over0$ evaluation, you can continue the process with $\frac{f''(0)}{g''(0)}$ and so on.
The direct method showed that terms were vanishing until the $x^3$ term, so it is expected that we have to go till $\frac{f'''(0)}{g'''(0)}$ to get a proper limit using L'Hospital.

$f(x)=\pi\sin(x)-\sin(\pi x)$ and $f(0)=0$
$f'(x)=\pi\cos(x)-\pi\cos(\pi x)$ and $f'(0)=0$
$f''(x)=-\pi\sin(x)+\pi^2\sin(\pi x)$ and $f''(0)=0$
$f'''(x)=-\pi\cos(x)+\pi^3\cos(\pi x)$ and $f'''(0)=\pi^3-\pi$

$g(x)=x\cos(x)-x\cos(\pi x)$ and $g(0)=0$
$g'(x)=\cos(x)-\cos(\pi x)-x\sin(x)+\pi x\sin(\pi x)$ and $g'(0)=0$
$g''(x)=-2\sin(x)+2\pi\sin(\pi x)-x\cos(x)+\pi^2x\cos(\pi x)$ and $g''(0)=0$
$g'''(x)=-3\cos(x)+3\pi^2\cos(\pi x)+x(...)$ and $g'''(0)=3\pi^2-3$
And we have the same $\frac{\pi}{3}$ limit than previously.
This method is teached first because it is an easy rule to apply, but in practice with ad hoc exercises like this one, it is tedious to derivate again and again. The Taylor expansion is generally more efficient in most cases, but in complex situations (with square roots, or composed functions), it becomes a bit technical to keep track of the right powers of $x$, you have to be consistent with the orders of $o(x^n)$ used, else your result will be wrong. 
A: As $x\to 0$, we get
$$f(x)=\pi\sin x-\sin \pi x\quad \to 0$$ $$g(x)=x(\cos x-\cos \pi x)\quad\to 0.$$ 
As $x\to 0$, we get 
$$f'(x)=\pi\cos x-\pi\cos \pi x\quad\to 0$$ $$g'(x)=x(-\sin x+\pi\sin\pi x)+(\cos x-\cos\pi x)\quad\to 0.$$ 
As $x\to 0$, we get
$$f''(x)=-\pi\sin x+\pi^2\sin\pi x\quad\to 0$$  
$$g''(x)=x(-\cos x+\pi^2\cos \pi x)+(-\sin x+\pi\sin \pi x)+(-\sin x+\pi\sin \pi x)$$ so $$g''(x)=x(-\cos x+\pi^2\cos \pi x)+2(-\sin x+\pi\sin \pi x)\quad\to 0.$$
As $x\to 0$, $$f'''(x)=-\pi\cos x+\pi^3\cos \pi x\quad\to -\pi+\pi^3$$  
$$g'''(x)=x(\sin x-\pi^3\sin\pi x)+(-\cos x+\pi^2\cos\pi x)+2(-\cos x+\pi^2\cos\pi x) $$ or
$$g'''(x)=x(\sin x-\pi^3\sin\pi x)+3(-\cos x+\pi^2\cos\pi x)\quad\to 3(-1+\pi^2).$$
Finally,
$$\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{f'(x)}{g'(x)}=\lim_{x\to 0}\frac{f''(x)}{g''(x)}=\lim_{x\to 0}\frac{f'''(x)}{g'''(x)}=\frac{-\pi+\pi^3}{3(-1+\pi^2)}=\frac{\pi(-1+\pi^2)}{3(-1+\pi^2)}=\frac{\pi}{3}.$$
