A) g'(x)
B) g(0)
C) g(0)+g'(x)
D) 0
Correct Answer: D
Solution :
[d] We have \[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+h)-(x)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x)+f(h)-f(x)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}g(h)}{h}=0.g(0)=0\] [\[\because \] g is continuous therefore \[\underset{h\to 0}{\mathop{\lim }}\,g(h)=g(0)\]]You need to login to perform this action.
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