A) ?1/9
B) ?2/9
C) ?1/3
D) 1/3
Correct Answer: B
Solution :
By definition, \[{f}'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{1}{2(1+h)-5}-\left( \frac{-1}{3} \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1}{2h-3}+\frac{1}{3} \right)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{3+2h-3}{3h(2h-3)} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{2h}{3h(2h-3)} \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2}{3(2h-3)}=\frac{2}{3(-3)}=\frac{-2}{9}\].
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