. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos (\sin x)-1}{{{x}^{2}}}=\] [Orissa JEE 2003]

. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos (\sin x)-1}{{{x}^{2}

1.	. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos (\sin x)-1}{{{x}^{2}}}=\] [Orissa JEE 2003]
A.	1
B.	? 1
C.	½
D.	?1/2
Answer» E.

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. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos (\sin x)-1}{{{x}^{2}}}=\] [Orissa JEE 2003]

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