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Question: \(\lim_{h \rightarrow 0}\frac{f\left( 2h + 2 + h^{2} \right) - f(2)}{f\left( h - h^{2} + 1 \right) -...

lim⁡h→0f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\lim_{h \rightarrow 0}\frac{f\left( 2h + 2 + h^{2} \right) - f(2)}{f\left( h - h^{2} + 1 \right) - f(1)} , given that f’(2) = 6 and f’(1) = 4

A

Does not exist

B

Is equal to −3/2

C

Is equal to 3/2

D

Is equal to 3

Answer

Is equal to 3

Explanation

Solution

lim⁡h→0f(2h+2+h2)−f(2)(h−h2+1)−f(1)=[00form]\lim_{h \rightarrow 0}\frac{f\left( 2h + 2 + h^{2} \right) - f(2)}{\left( h - h^{2} + 1 \right) - f(1)} = \left\lbrack \frac{0}{0}form \right\rbrack

∴ Applying L’ Hospital’s rule we get

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