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$$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$ is equal to :

JEE · Math · previous-year question

  1. A.$$ - \pi $$
  2. B.$$ \pi $$correct
  3. C.$${\pi \over 2}$$
  4. D.1

Answer

B. $$ \pi $$

Explanation

Consider $$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$ $$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]} \over {{x^2}}}$$ $$ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi - \pi {{\sin }^2}x} \right)} \over {{x^2}}}$$ $$\left[ \, \right.$$ As $$\sin \left( {\pi - \theta } \right) = \sin \theta $$ $$\left. \, \right]$$ $$ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}} \times {{\pi {{\sin }^2}x} \over {{x^2}}}$$ $$ = \mathop {\lim }\limits_{x \to 0} 1 \times \pi {\left( {{{\sin x} \over x}} \right)^2} = \pi $$

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