Mathematics - Limits Question with Solution | TestHub

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

$=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }-{\mathrm{\pi cos}}^2\mathrm{x}\right)}{{\mathrm{x}}^2}$$\left[\because \text{sin}\left(\pi -\theta \right)=\text{sin }\theta \right]$

$\text{=}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\mathrm{\pi } {\sin }^2\mathrm{x}\right)}{\left(\mathrm{\pi } {\sin }^2\mathrm{x}\right)}\times \frac{\mathrm{\pi } {\sin }^2\mathrm{x}}{{\mathrm{x}}^2}$

$\text{=}\underset{\mathrm{x}\to 0}{\lim }\mathrm{\pi }\left[\frac{\sin \left(\mathrm{\pi } {\sin }^2\mathrm{x}\right)}{\left(\mathrm{\pi } {\sin }^2\mathrm{x}\right)}\right]$${\left[\frac{\text{sin }\text{x}}{\text{x}}\right]}^2$

$=\pi \left(1\right)\left(1\right)$$\left[\mathrm{As}, \underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{t}}{\mathrm{t}}=1\right]$

$=\pi$.