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