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 $

Answer

Correct option: D.
$\pi $
d
Consider

$\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$

$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi  - \pi {{\sin }^2}x} \right)}}{{{x^2}}}$

              [$\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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