1708\hspace{-16}$ Given $\bf{\lim_{x\rightarrow 0}\frac{\sin (\pi \cos^2 x)}{x^2}}$\\\\\\ $\bf{\Rightarrow \lim_{x\rightarrow 0}\frac{\sin (\pi-\pi\sin^2 x)}{x^2}}$\\\\\\ $\bf{\Rightarrow \lim_{x\rightarrow 0}\left\{\frac{\sin (\pi\sin^2 x)}{(\pi \sin^2 x)}\times \frac{\sin^2 x}{x^2}\right\}\times \pi = \pi}$\\\\\\ Bcz $\bf{\lim_{x\rightarrow 0}\frac{\sin (\pi\sin^2 x)}{(\pi\sin^2 x)} = 1}$ and $\bf{\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1}$