MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}\) is equal to
- A \(-\pi\)
- B \(\pi\)
- C \(\frac{\pi}{2}\)
- D 1
Answer & Solution
Correct Answer
(B) \(\pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{\substack{x \rightarrow 0}} \frac{\sin \left(\pi\left\{1-\sin ^2 x\right\}\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2}=\lim _{\substack{2}} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \cdot \frac{\pi \sin ^2 x}{x^2}=\pi\end{aligned}\)
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