\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{e}^{x}}-\log (1+x)}{{{x}^{2}}}\] equals [RPET 1996]

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

1.	\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{e}^{x}}-\log (1+x)}{{{x}^{2}}}\] equals [RPET 1996]
A.	\[\frac{2}{3}\]
B.	\[\frac{1}{3}\]
C.	\[\frac{1}{2}\]
D.	\[\frac{3}{2}\]
Answer» E.

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\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{e}^{x}}-\log (1+x)}{{{x}^{2}}}\] equals [RPET 1996]

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