A) \[-2\]
B) \[-3\]
C) \[1\]
D) does not exist
Correct Answer: B
Solution :
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g(f(x))\] \[=g\,(f({{0}^{+}}))=g\,({{1}^{+}})\] \[\left( \because \,\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}={{1}^{+}} \right)\] \[=1-2\,(1)-2=-\,3\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(f(x)=g\,(f({{0}^{-}}))=g({{2}^{+}})\left( \because \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,(2-x)={{2}^{+}} \right)\]\[=2-5=-\,3\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,g(f(x))=-\,3\]You need to login to perform this action.
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