A) f is continuous at x=1
B) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=log3\]
C) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=-\sin 1\]
D) \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\]does not exist
Correct Answer: C
Solution :
[c] For \[\left| x \right|<1,{{x}^{2n}}\to 0\] as \[n\to \infty \] and for \[\left| x \right|>1,\,\,1/{{x}^{2n}}\to 0\] as \[n\to \infty ,\]So, \[f(x)=\left\{ \begin{matrix} \log (2+x)\, \\ \underset{\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-2n}}\log (2+x)-sinx}{{{x}^{-2n}}+1}=-\sin x, \\ \frac{1}{2}[log(2+x)-sinx],\, \\ \end{matrix} \right.\,\,\,\begin{matrix} \left| x \right|<1 \\ \left| x \right|>1 \\ \left| x \right|=1 \\ \end{matrix}\]Thus, \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(-\sin x)=-sin1\] and \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,\,\,log(2+x)=\log 3.\]
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