A) \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\ne 2\]
B) \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=0\]
C) \[f(x)\]is continuous at \[x=0\]
D) none of the above
Correct Answer: C
Solution :
We have,\[f(x)=\left\{ \begin{matrix} \frac{\sin x}{x}+\cos x, & x\ne 0 \\ 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,, & x=0 \\ \end{matrix} \right.\] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sin x}{x}+\cos x\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (0-h)}{0-h}+\cos (0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{h}+\cos \,h\] \[=1+1=2\] \[RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sin x}{x}+\cos x\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (0+h)}{h}+\cos (0+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{h}+\cos h\] \[=1+1=2\] \[\therefore \] \[LHL=RHL=f(0)=2\] \[\Rightarrow \]\[f(x)\] is continuous at \[x=0\]
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