<s>[INST] $ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$ I have to evaluate the following limit
$$ \lim_{x \to  0}{\frac{\sin( \pi \cos x)}{x \sin x} }$$
My solution is:
$$ \lim_{x \to  0}{\frac{\sin( \pi \cos x)}{x \sin x} }=\lim_{x \to  0}{\frac{\pi \cos x}{x \cdot  x} }=+\infty$$
But the correct result is $\frac{\pi}{2}$. I can't understand where I'm making mistakes.
 [/INST] By some elementary inequalities, near $x=0$ we have
$$
\cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have
$$
\frac{\sin(\pi\cos(x))}{x\sin(x)} = \frac{\sin(\pi(1-x^2/2)+ O(x^4))}{x\sin(x)}
$$Use the sine-reflection and angle-addition:
$$
 \frac{\sin(\pi(1-x^2/2)+O(x^4))}{x\sin(x)} = \frac{\sin(\pi/2 \cdot x^2+O(x^4))}{x\sin(x)}
$$
$$
= \frac{\sin(\pi/2\cdot x^2)\cos(O(x^4))}{x\sin(x)}+ \frac{\cos(\pi/2 \cdot x^2)\sin(O(x^4))}{x\sin(x)}
$$Using $\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}=1$, the second term vanishes in the limit, and $\cos(O(x^4))\to 1$. So we have
$$
\lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)\cdot 1}{x\sin(x)} = \lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)}{x^2}\cdot \frac{x}{\sin(x)} =\frac{\pi}{2}\cdot 1 =\frac{\pi}{2}
$$
</s>