Solve
Q. $\underset{x\to 0}{\lim }\frac{\sin \left({\mathrm{\pi cos}}^2\mathrm{x}\right)}{x^2}$

$$\begin{gathered} Here property used for first is \\ \underset{x\to 0}{\lim }\frac{\sin x}{x}= 1 \\ and for second \\ \underset{x\to 0}{\lim }\frac{{\mathrm{\pi sin}}^{2}x}{x^2} it is 0/0 form so u\sin g L\text{'}hospital rule \\ = \underset{x\to 0}{\lim }\frac{\pi {\displaystyle \frac{d}{dx}}{\sin }^{2}x}{\frac{d}{dx}{x}^2}=\underset{x\to 0}{\lim }\frac{2\pi \sin x\cos x}{2x} \\ =\pi \underset{x\to 0}{\lim }\cos x\times \underset{x\to 0}{\lim }\frac{\sin x}{x}= \pi \end{gathered}$$