A) \[0\]
B) \[2\]
C) \[sin\text{ }1\]
D) None of these
Correct Answer: B
Solution :
Given, \[f(x)=\left\{ \begin{align} & \frac{1}{2}\{g(x)+(x)\}\,\sin \,x,\,x\ge 1 \\ & \frac{\sin \,x}{x},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x<1 \\ \end{align} \right.\] and \[g(x)=\left\{ \begin{matrix} 1, & if & x>0 \\ -1, & if & x<0 \\ 0, & if & x=0 \\ \end{matrix} \right.\] Now, \[RHL=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{1}{2}\{g(x)+(x)\}\,\sin x\] \[=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{1}{2}\{1+x)\,\sin x\] \[=\frac{1}{2}.(1+1)\,sin1\] \[=\,sin1\] and \[LHL=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sin \,x}{x}\] \[=\frac{\sin \,(-1)}{(-1)}\] \[=\frac{-\sin \,1}{-1}\] \[=\sin 1\] Since, \[RHL=LHL=\sin \,1\] \[\therefore \] \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\sin 1\]You need to login to perform this action.
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