A) 1
B) \[-1\]
C) 0
D) does not exist
Correct Answer: D
Solution :
We have \[f(x)=\left\{ \begin{matrix} \frac{|x-4|}{x-4}, & x\ne 4 \\ 0, & x=4 \\ \end{matrix} \right.\] Now, RHL \[\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(4+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|4+h-4|}{4+h-4}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,1=1\] and LHL \[\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(4-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|4-h+4|}{4-h+4}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-h|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,(-1)=-1\] Since, \[LHL\ne RHL\] ie, \[\underset{x\to {{4}^{-1}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)\] Thus, limit does not exist.
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