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Question: If f ¢(3) = 2 then \(\lim_{h \rightarrow 0}\frac{f(3 + h^{2}) - f(3 - h^{2})}{2h^{2}} =\)...

If f ¢(3) = 2 then lim⁡h→0f(3+h2)−f(3−h2)2h2=\lim_{h \rightarrow 0}\frac{f(3 + h^{2}) - f(3 - h^{2})}{2h^{2}} =

A

1

B

2

C

3

D

½

Answer

2

Explanation

Solution

lim⁡h→0f(3+h2)−f(3−h2)2h2(00form)\lim_{h \rightarrow 0}\frac{f(3 + h^{2}) - f(3 - h^{2})}{2h^{2}}\left( \frac{0}{0}form \right)

By L ¢ Hospital lim⁡h→02hf′(3+h2)+2hf′(3−h2)4h\lim_{h \rightarrow 0}\frac{2hf'(3 + h^{2}) + 2hf'(3 - h^{2})}{4h}

= lim⁡h→0f′(3+h2)+f′(3−h2)2\lim_{h \rightarrow 0}\frac{f'(3 + h^{2}) + f'(3 - h^{2})}{2} = f′(3)+f′(3)2=2f′(3)2\frac{f'(3) + f'(3)}{2} = \frac{2f'(3)}{2}

Ž f ¢(3) = 2