\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x+\log (1-x)}{{{x}^{2}}}\] is equal to [Roorkee 1995]

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

1.	\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x+\log (1-x)}{{{x}^{2}}}\] is equal to [Roorkee 1995]
A.	0
B.	\[\frac{1}{2}\]
C.	\[-\frac{1}{2}\]
D.	None of these
Answer» D. None of these

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\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x+\log (1-x)}{{{x}^{2}}}\] is equal to [Roorkee 1995]

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