| Statement 1: Let \[f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{\text{In}(1+x)-{{x}^{2n}}(\sin 2x)}{1+{{x}^{2n}}}\], then \[f(x)\] is discontinuous at \[x=1\]. |
| Statement 2: \[LHL=RHL\ne f(1)\] |
A) Both statements are true, and Statement-2 explains Statement-1.
B) Both statements are true, but Statement-2 does not explain Statement-1.
C) Statement-1 is True, Statement-2 is False.
D) Statement-1 is False, Statement-2 is true.
Correct Answer: C
Solution :
\[\because \,\,\underset{n\to \infty }{\mathop{\lim }}\,\,{{x}^{2n}}=\left\{ \begin{matrix} 0, & 0<x<1 \\ \begin{align} & 1, \\ & \infty , \\ \end{align} & \begin{align} & x=1 \\ & x>1 \\ \end{align} \\ \end{matrix} \right.\] \[\because \,\,LHL=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left\{ \underset{n\to \infty }{\mathop{\lim }}\,\frac{In(1+x)-{{x}^{2n}}\sin 2x}{1+{{x}^{2n}}} \right\}=In\,2\] \[\because \,\,RHL=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left\{ \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-2n}}In(1+x)-\sin 2x}{{{x}^{-2n}}+1} \right\}=-\sin 2\]\[f(1)=\frac{In\,2-\sin 2}{2}\] \[\therefore \,\,LHL\ne RHL\ne f(1)\]
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