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Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    Let\[f(x)=\left\{ \begin{matrix}    \sin x, & x\ne n\pi   \\    2 & x=n\pi   \\ \end{matrix} \right.\], where\[n\in I\]and\[g(x)=\left\{ \begin{matrix}    {{x}^{2}}+1, & x\ne 2  \\    3, & x=2  \\ \end{matrix} \right.\], then\[\underset{x\to 0}{\mathop{\lim }}\,g[f(x)]\]is

    A)  1                                            

    B)  0

    C)  3                            

    D)         does not exist

    Correct Answer: A

    Solution :

                    \[g[f(x)]=\left\{ \begin{matrix}    {{[f(x)]}^{2}}+1, & f(x)\ne 2  \\    3, & f(x)=2  \\ \end{matrix} \right.\] \[\Rightarrow \]               \[g[f(x)]=\left\{ \begin{matrix}    {{\sin }^{2}}x+1, & x\ne n\pi   \\    3, & x=n\pi   \\ \end{matrix} \right.\] \[\therefore \]  \[RHL=\underset{h\to 0}{\mathop{\lim }}\,g[f(0+h)]\]                 \[=\underset{h\to 0}{\mathop{\lim }}\,({{\sin }^{2}}h+1)=1\] and        \[LHL=\underset{h\to 0}{\mathop{\lim }}\,[f(0-h)]\]                 \[=\underset{h\to 0}{\mathop{\lim }}\,({{\sin }^{2}}h+1)=1\] \[\therefore \]  \[\underset{x\to 0}{\mathop{\lim }}\,g[f(x)]=1\]


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