Let's do some calculus! (3)

We note that as $x \to 0$ , $\sin(\pi \cos^2 x) \to 0$ and $x^2 \to 0$ . So this is a 0/0 case and we can use the L'Hôpital's rule as follows.

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin(\pi \cos^2 x)}{x^2} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t. }x} \\ & = \lim_{x \to 0} \frac {\cos (\pi \cos^2 x) (-2 \pi \cos x \sin x)}{2x} & \small \color{#3D99F6}{\text{Still 0/0, differentiate again}} \\ & = \lim_{x \to 0} \frac {- \sin (\pi \cos^2 x) (\pi^2 \color{#D61F06}{\sin^2 2x}) + \cos (\pi \color{#3D99F6}{\cos^2 x}) (-2\pi \color{#3D99F6}{\cos 2 x})}{2} \\ & = \frac {\color{#D61F06}{0} + \cos (\pi (\color{#3D99F6}{1})) (-2\pi (\color{#3D99F6}{1}))}2 = \boxed{\pi} \end{aligned}$