A) 3
B) \[-\frac{3}{2}\]
C) \[\frac{3}{2}\]
D) Does not exist
Correct Answer: A
Solution :
\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\{f'(2h+2+{{h}^{2}})\}(2+2h)-0}{\{f'(h-{{h}^{2}}+1)\}(1-2h)-0}\] (using L Hospital?s rule) \[=\frac{f'(2).2}{f'(1).1}=\frac{6\times 2}{4\times 1}\] \[=\frac{12}{4}=3\]You need to login to perform this action.
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