Answer:
At x = 0, \[LHL=\underset{x\to \,\,{{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to \,\,0}{\mathop{\lim }}\,f(0-h)\] \[=\underset{x\to \,\,{{0}^{-}}}{\mathop{\lim }}\,f(-h)=\underset{h\to \,\,0}{\mathop{\lim }}\,-h\sin \left( \frac{1}{-\,h} \right)\] \[=0\times \] (An oscillating number between \[-\,1\] and 1) = 0 RHL \[=\underset{x\to \,\,{{0}^{+}}}{\mathop{\lim }}\,f(x)\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,f(0+h)=\underset{h\to \,\,0}{\mathop{\lim }}\,f(h)\] \[=\underset{h\to \,0}{\mathop{\lim }}\,h\,\,\sin \frac{1}{h}\] \[=0\times \] (An oscillating number between \[-\,1\] and 1) = 0 and f(0) = 0 Thus f(0) = LHL = RHL \[\therefore \] f(x) is continuous at x = 0.
You need to login to perform this action.
You will be redirected in
3 sec
