Let\[f(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{\log (2+x)-{{x}^{2n}}\sin x}{1+{{x}^{2n}}}\]. Then

\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=sin1\]

\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\]

\[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\] doesn?t exist

Let\[f(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{\log (2+x

1.	Let\[f(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{\log (2+x)-{{x}^{2n}}\sin x}{1+{{x}^{2n}}}\]. Then
A.	\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\]
B.	\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=sin1\]
C.	\[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\] doesn?t exist
D.	None of these
Answer» B. \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=sin1\]

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Let\[f(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{\log (2+x)-{{x}^{2n}}\sin x}{1+{{x}^{2n}}}\]. Then

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