Both the numerator and denominator approach 0, so by L'Hopital's rule
[tex]\displaystyle\lim_{x\to0}\frac{\sin(\pi\cos x)}{x^2}=\lim_{x\to0}\frac{-\pi\sin x\cos(\pi\cos x)}{2x}[/tex]
Using the fact that
[tex]\displaystyle\lim_{x\to0}\frac{\sin x}x=1[/tex]
we find
[tex]\displaystyle\lim_{x\to0}\frac{-\pi\sin x\cos(\pi\cos x)}{2x}=\lim_{x\to0}\frac{-\pi\cos(\pi\cos x)}2[/tex]
[tex]\pi\cos x\to\pi[/tex] as [tex]x\to0[/tex], and [tex]\cos\pi=-1[/tex], so the limit is π/2.
(I'm wondering if there's a way to do this without L'Hopital's rule, in case you haven't learned it yet...)
