A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 2
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2\,{{\sin }^{2}}\frac{x}{2}.\,({{x}^{2}})}{4\,{{\sin }^{2}}x\,.\,\left( \frac{{{x}^{2}}}{4} \right)}=\frac{1}{2}.\] Aliter : Apply L-Hospital?s rule two times.
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