A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[-\pi \]
D) \[1\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin \,(\pi \,{{\cos }^{2}}x)}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,(\pi {{\cos }^{2}}x)}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,(\pi \,{{\cos }^{2}}x)}{\pi \,{{\cos }^{2}}x}\times \frac{\pi \,{{\cos }^{2}}x}{{{x}^{2}}}\] \[=1\times \pi \times 1=\pi \] \[\left( \because \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin x}{x}=1\,and\,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\cos \,x}{x}=1 \right)\]
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