Question

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

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

Answer 1

The correct option is D

$\mathrm{\pi }$

Explanation for the correct option:

Find the value of $\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }{\cos }^2\mathrm{x}\right)}{x^2}$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }{\cos }^2\mathrm{x}\right)}{x^2} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }\left(1-{\sin }^2\mathrm{x}\right)\right)}{x^2}\mathbf{[}\mathbf{\text{Where,}}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{]} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }-\mathrm{\pi }{\sin }^2\mathrm{x}\right)}{x^2} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }{\sin }^2\mathrm{x}\right)}{x^2} \end{gathered}$$

Multiply and divided by $\mathrm{\pi }{\sin }^2\mathrm{x}$

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

We know that

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{}\mathbf{x}}{\mathbf{x}}\mathbf{=}\mathbf{1}$

Then,

$$\begin{gathered} I=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }{\sin }^2\mathrm{x}\right)}{\mathrm{\pi }{\sin }^2\mathrm{x}}\times \frac{\mathrm{\pi }{\sin }^2\mathrm{x}}{{\mathrm{x}}^2} \\ \Rightarrow I=1.\mathrm{\pi }\left(1\right) \\ \Rightarrow I=\mathrm{\pi } \end{gathered}$$

Then,

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

Hence, the correct answer is option D.