A) continuous but non-differentiable at x=0
B) differentiable at x=0
C) discontinuous at x=0
D) none of these
Correct Answer: C
Solution :
[c] \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,f(h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{2{{h}^{2}}+h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{2h+1}=1\] And \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,f(-h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h}{2{{h}^{2}}+\left| -h \right|}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h}{2{{h}^{2}}+h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-1}{2h+1}=-1\]
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