If fx=9−x2 then Ltx→2fx−f2x−2 is

Givenf(x)=9−x2⇒f′(x)=−2x29−x2=−x9−x2Now1tf(x)−f(2)x−2=f′(2)=−25

If fx=9−x2 then Ltx→2fx−f2x−2 is

Givenf(x)=9−x2⇒f′(x)=−2x29−x2=−x9−x2Now1tf(x)−f(2)x−2=f′(2)=−25

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If $f (x) = \sqrt{9 - x^{2}}$ then $\underset{x \to 2}{L t} \frac{f (x) - f (2)}{x - 2}$ is

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$0$

$- 2 / \sqrt{5}$

$3 / \sqrt{5}$

$4$

answer is B.

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Detailed Solution

$Given f (x) = \sqrt{9 - x^{2}} \Rightarrow f^{'} (x) = \frac{- 2 x}{2 \sqrt{9 - x^{2}}} = \frac{- x}{\sqrt{9 - x^{2}}}$
$Now 1 t \frac{f (x) - f (2)}{x - 2} = f^{'} (2)$
$= \frac{- 2}{\sqrt{5}}$

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If $f (x) = \sqrt{9 - x^{2}}$ then $\underset{x \to 2}{L t} \frac{f (x) - f (2)}{x - 2}$ is