A) \[-\frac{1}{9}\]
B) \[-\frac{2}{9}\]
C) \[-13\]
D) \[\frac{1}{3}\]
E) None of these
Correct Answer: B
Solution :
\[f(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{1+h-1}{2{{(1+h)}^{2}}-7(1+h)+5}-\left( -\frac{1}{3} \right)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{h}{2(1+{{h}^{2}}+2h)-7-7h+5}+\frac{1}{3}}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{h}{h(2h-3)}+\frac{1}{3}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1}{2h-3}+\frac{1}{3} \right)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{3+2h-3}{3h(2h-3)} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{2h}{3h(2h-3)} \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2}{3(2h-3)}=\frac{2}{3(-3)}=-\frac{2}{9}\]
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