A) \[g'(x)\]
B) \[g(0)\]
C) \[g(0)+g'(x)\]
D) 0
Correct Answer: D
Solution :
We have \[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+h)-f(x)}{h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x)+f(h)-f(x)}{h}\] \[\left[ \because f(x+y)=f(x)+f(y) \right]\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}g(h)}{h}=0.g(0)=0\] \[[\because \,\,g\]is continuous therefore \[\underset{h\to 0}{\mathop{\lim }}\,g(h)=g(0)]\].
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