Solution
= Solution
Both numerator and denominator have a double zero at one. Applying <L'Hôpital's rule> twice gives
$$
\boxed{\lim_{x\to1}\frac{1-\sin(\pi x/2)}{x^3-3x+2}=\lim_{x\to1}\frac{-(\pi/2)\cos(\pi x/2)}{3x^2-3}=\lim_{x\to1}\frac{(\pi^2/4)\sin(\pi x/2)}{6x}=\frac{\pi^2}{24}.}
$$
Alternatively $x^3-3x+2=(x-1)^2(x+2)$ and $1-\sin(\pi x/2)\sim\pi^2(x-1)^2/8$, giving the same result from a <Taylor expansion>.