MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \,\left( {\pi {{\cos }^2}\,x} \right)}}{{{x^2}}}$ equals
- A$-\pi $
- B$1$
- C$-1$
- ✓$\pi $
$\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$
$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi {{\sin }^2}x} \right)}}{{{x^2}}}$
[$\because $ $\sin \left( {\pi - \theta } \right) = \sin \theta $]
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \times \frac{{\left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}} = \pi $
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