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
Correct Answer: A
Solution :
| [a] For \[\left| x \right|<1,{{x}^{2n}}\to 0\] as \[n\to \infty \] and |
| For \[\left| x \right|>1,\frac{1}{{{x}^{2n}}}\to 0\] as \[n\to \infty \]. |
| So, |
| \[f(x)=\left\{ \begin{matrix} \log (2+x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\left| x \right|<1 \\ \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-2n}}\log \,(2+x)-sin\,\,x}{{{x}^{-2n}}+1}=-\sin x\,\,if\,\left| x \right|>1 \\ \frac{1}{2}[log\,(2+x)-sin\,\,x]if\left| x \right|=1 \\ \end{matrix} \right.\] |
| Thus \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x{{\to }^{1+}}}{\mathop{\lim }}\,(-sinx)=-sin1\] |
| and \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x{{\to }^{1-}}}{\mathop{\lim }}\,\log (2+x)=log3.\] |
| Hence, L.H.L\[\ne \]R.H.L |
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