Question

${lim}_{x\to 0}\frac{\sin \left(\pi {\cos }^2\pi \right)}{x^2}$ is equal to

• A. $\frac{\pi }{2}$
• B. $1$
• C. $-\pi$
• D. $\pi$

Answer 1

The correct option is D $\pi$

${lim}_{x\to 0}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}$

As ${\sin }^{2}x+{\cos }^{2}x=1\Rightarrow {lim}_{x\to 0}\frac{\sin \left(\pi (1-{\sin }^{2}x)\right)}{x^2}$

$={lim}_{x\to 0}\frac{\sin (\pi -\pi {\sin }^{2}x)}{x^2}={lim}_{x\to 0}\frac{\sin \left(\pi {\sin }^{2}x\right)}{x^2}\left[{lim}_{x\to 0}\frac{\sin \theta }{\theta }=1\right]$

$={lim}_{x\to 0}\frac{\pi {\sin }^{2}x}{x^2}={lim}_{x\to 0}{\left(\frac{\sin x}{x}\right)}^2=\pi$