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  4. If f'(2)=6,f'(1)=4, , then underseth→ 0 mathop lim (f(2h+2+h2)-f(2)/f(h-h2+1)-f(1)) is equal to

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Q. If $ f'(2)=6,f'(1)=4, $ , then $ \underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)} $ is equal to

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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 $