A) 0
B) 1/2
C) \[-1\]
D) \[-1/2\]
E) none of these
Correct Answer: A
Solution :
\[f(x)=|x{{|}^{3}}=\left\{ \begin{matrix} 0, & x=0 \\ {{x}^{3}}, & x>0 \\ -{{x}^{3}}, & x<0 \\ \end{matrix} \right.\] Now, \[Rf(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}-0}{h}=0\] and \[Lf(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(-h)-f(0)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-{{h}^{3}}-0}{-h}=0\] \[\because \] \[Rf(0)=Lf(0)=0\] \[\therefore \] \[f(0)=0\]
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