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Question: Given that \(f'(2) = 6\) and \(f^{'}(1) = 4\), then\lim_{h \rightarrow 0}\frac{f(2h + 2 + h^{2}) - f...

Given that f′(2)=6f'(2) = 6 and f′(1)=4f^{'}(1) = 4, then\lim_{h \rightarrow 0}\frac{f(2h + 2 + h^{2}) - f(2)}{f(h - h^{2} + 1) - f(1)} =

A

Does not exist

B

−32- \frac{3}{2}

C

32\frac{3}{2}

D

3

Answer

3

Explanation

Solution

lim⁡h→0f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\lim_{h \rightarrow 0}\frac{f(2h + 2 + h^{2}) - f(2)}{f(h - h^{2} + 1) - f(1)}

=lim⁡h→0f′(2h+2+h2)(2+2h)f′(h−h2+1)(1−2h)=6×24×1=3= \lim_{h \rightarrow 0}\frac{f^{'}(2h + 2 + h^{2})(2 + 2h)}{f^{'}(h - h^{2} + 1)(1 - 2h)} = \frac{6 \times 2}{4 \times 1} = 3.