A) 0
B) 1
C) 2
D) Does not exist
Correct Answer: D
Solution :
\[R{f}'(2)\]\[=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{f(2+h)-f(2)}{h}\] \[=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{2(2+h)-1-(4-1)}{h}\]\[=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{4+2h-1-3}{h}=2\] and \[L{f}'(2)=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{f(2-h)-f(2)}{-h}=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{2-h+1-3}{-h}=1\]. Thus \[{f}'(2)\] does not exist.
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