MCQ
$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{\rm{sin}}\left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = $
- A$ - \pi $
- B$\;\pi $
- C$\frac{\pi }{2}$
- D$1$
$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi \left( {1 - {{\sin }^2}x} \right)} \right)}}{{{x^2}}}$
$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$
$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$
$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \cdot \frac{{\pi {{\sin }^2}x}}{{{x^2}}}$
We know, $\mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {f\left( x \right)} \right)}}{{f\left( x \right)}} = 1$
So, our limits becomes,
$ = 1 \cdot \pi \left( 1 \right) = \pi $
$\therefore \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = \pi $
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