A) \[f(x)g(0)\]
B) \[3g(0)\]
C) \[f(x)\cos 3x\]
D) \[3f(x)g(0)\]
E) \[3f(x)g(x)\]
Correct Answer: D
Solution :
\[f(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+h)-f(x)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x)f(h)-f(x)}{h}\] \[=f(x)\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-1}{h}\] \[=f(x)\underset{h\to 0}{\mathop{\lim }}\,\frac{1+\sin 3h\,g(h)-1}{h}\] \[=3f(x)\left( \underset{h\to 0}{\mathop{\lim }}\,\frac{\sin 3h}{3h} \right)\left( \underset{h\to 0}{\mathop{\lim }}\,g(h) \right)\] \[=3f(x)g(0)\]
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