Question
$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \,{{\cos }^2}\,x} \right)}}{{{x^2}}}$$ is equal to-
A.
$$ - \pi $$
B.
$$\pi $$
C.
$$\frac{\pi }{2}$$
D.
$$1$$
Answer :
$$\pi $$
Solution :
Consider
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \,{{\cos }^2}\,x} \right)}}{{{x^2}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]}}{{{x^2}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \,\sin \frac{{\left( {\pi - \pi {{\sin }^2}x} \right)}}{{{x^2}}}\,\,\,\,\,\,\,\,\,\left[ {\because \sin \,\left( {\pi - \theta } \right) = \sin \,\theta } \right] \cr
& = \mathop {\lim }\limits_{x \to 0} \,\sin \frac{{\left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \times \frac{{\pi {{\sin }^2}x}}{{{x^2}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \,1 \times \pi {\left( {\frac{{\sin x}}{x}} \right)^2} \cr
& = \pi \cr} $$