A) \[-1\]
B) \[-0.5\]
C) \[0.5\]
D) \[1\]
Correct Answer: C
Solution :
Given, \[f(x)={{x}^{3}}\] \[\therefore \] \[f(x+h)={{(x+h)}^{3}}\] Now, \[f'(x)=3{{x}^{2}}\] \[\therefore \] \[f'(x+\theta h)=3{{(x+\theta h)}^{2}}\] Given, \[\frac{f(x+h)-f(x)}{h}=f'(x+\theta h)\] \[\Rightarrow \] \[\frac{{{(x+h)}^{3}}-{{x}^{3}}}{h}=3{{(x+\theta h)}^{2}}\] \[\Rightarrow \] \[\frac{{{x}^{3}}+{{h}^{3}}+3xh(x+h)-{{x}^{3}}}{h}\] \[=3({{x}^{2}}+{{\theta }^{2}}{{h}^{2}}+2x\theta h)\] \[\Rightarrow \] \[{{h}^{2}}+3{{x}^{2}}+3xh=3{{x}^{2}}+3{{\theta }^{2}}{{h}^{2}}+6x\theta h\] \[\Rightarrow \] \[3x=3{{\theta }^{2}}h+6x\,\theta \] Taking limit on both sides, we get \[\underset{h\to 0}{\mathop{\lim }}\,\,(h+3x)=\underset{h\to 0}{\mathop{\lim }}\,(3{{\theta }^{2}}h+6x\theta )\] \[\Rightarrow \] \[3x=6\,\,\underset{h\to 0}{\mathop{\lim }}\,\theta \] \[\Rightarrow \] \[\underset{h\to 0}{\mathop{\lim }}\,\theta =\frac{1}{2}=0.5\]
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