$ lim_{x to 0}{frac{sin( pi cos x)}{x sin x} }$

$$L=lim_{x to 0}{frac{sin( pi cos x)}{x sin x} }$$ $$L=lim_{x to 0}{frac{sin( pi cos x)}{x^2} }$$ By L'Hôpital's rule: $$L=-lim_{x to 0}{frac{cos( pi cos x)pi sin x}{2x} }$$ $$L=-lim_{x to 0}{frac{cos( pi cos x)pi }{2} }$$ $$implies L=dfrac {pi}2$$

Only using

1. well know limits,
2. the identity $sintheta=sin(pi-theta)$
3. the fact that $pi(1-cos{x})to0$ when $xto0$: begin{align} &lim_{xto0}{frac{sin(picos{x})}{xsin{x}}}=\ &=lim_{xto0}{frac{sin(pi-picos{x})}{xsin{x}}}=\ &=lim_{xto0}{frac{sin[pi(1-cos{x})]}{pi(1-cos{x})}cdotfrac{pi(1-cos{x})}{xsin{x}}}=\ &=lim_{xto0}{frac{sin[pi(1-cos{x})]}{pi(1-cos{x})}cdotpicdotfrac{1-cos{x}}{x^2}cdotfrac{x}{sin{x}}}=\ &=1cdotpicdotfrac{1}{2}cdot1=frac{pi}{2}\ end{align}

Use the identity $>picos x=pibigl(1-2sin^2{xover2}bigr)$ to obtain $${sin(picos x)over xsin x}={sinbigl(2pisin^2{xover2}bigr)over xcdot2sin{xover2}cos{xover2}}={piover2}cdot{sinbigl(2pisin^2{xover2}bigr)over2pisin^2{xover2}}cdot{tan{xover2}over{xover2}} .$$ On the right hand side the two last factors tend to $1$ when $xto0$. It follows that $$lim_{xto0}{sin(picos x)over xsin x}={piover2} .$$

By some elementary inequalities, near $x=0$ we have $$ cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have $$ frac{sin(picos(x))}{xsin(x)} = frac{sin(pi(1-x^2/2)+ O(x^4))}{xsin(x)} $$Use the sine-reflection and angle-addition: $$ frac{sin(pi(1-x^2/2)+O(x^4))}{xsin(x)} = frac{sin(pi/2 cdot x^2+O(x^4))}{xsin(x)} $$ $$ = frac{sin(pi/2cdot x^2)cos(O(x^4))}{xsin(x)}+ frac{cos(pi/2 cdot x^2)sin(O(x^4))}{xsin(x)} $$Using $lim_{thetato 0}frac{sin(theta)}{theta}=1$, the second term vanishes in the limit, and $cos(O(x^4))to 1$. So we have $$ lim_{xto 0}frac{sin(pi/2 cdot x^2)cdot 1}{xsin(x)} = lim_{xto 0}frac{sin(pi/2 cdot x^2)}{x^2}cdot frac{x}{sin(x)} =frac{pi}{2}cdot 1 =frac{pi}{2} $$

There are many ways in which you can calculate this. For example: L'Hospital's rule applied once will get you to

$$underset{xto 0}{text{lim}}-frac{pi (x sin ) (cos (pi (x cos )))}{x (cos x)+sin x}$$

(I leave the derivatives to you, as a warm up).

Now by product rule you can split the fraction as a product:

$$-pi left(underset{xto 0}{text{lim}}cos (pi (cos x))right) underset{xto 0}{text{lim}}frac{sin x}{x (cos x)+sin x}$$

Easily observing that the brackets limit is $-1$ which then makes $-pi$ to $pi$.

Now you can can multiply and divide by $x$ to get

$$pi lim_{xto 0} frac{color{red}{sin(x)}}{xcos(x) + sin(x)}cdot frac{x}{color{red}{x}}$$

The red zone goes to $1$. Collect $x$ over and above:

$$pi lim_{xto 0} frac{x}{xleft(cos(x) + frac{color{red}{sin(x)}}{color{red}{x}}right)}$$

Again a red zone that goes to $1$.

$x$ can be simplified. The cosine of $xto 0$ goes to $1$ leading to

$$frac{pi}{1+1} = frac{pi}{2}$$