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If $\mathrm{F}(\mathrm{x})=\sqrt{9-\mathrm{x}^{2}}$, then what is $\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{F}(\mathrm{x})-\mathrm{F}(1)}{\mathrm{x}-1}$ equal to?
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The correct answer is:
$-\frac{1}{2 \sqrt{2}}$
$\lim _{x \rightarrow 1} \frac{F(x)-F(1)}{x-1}$
$\quad$ We know, $_{x \rightarrow a}{\lim } \frac{F(n)-F(a)}{x-a}=F^{\prime}(a)$
$\therefore F^{\prime}(x)=\frac{d}{d x}\left(\sqrt{9-x^{2}}\right)$
$=\frac{1(0-2 x)}{2 \sqrt{9-x^{2}}}=\frac{-x}{\sqrt{9-x^{2}}}$
$\quad F^{\prime}(1)=\frac{-1}{\sqrt{9-1}}=\frac{-1}{\sqrt{8}}=\frac{-1}{2 \sqrt{2}}$
$\quad$ We know, $_{x \rightarrow a}{\lim } \frac{F(n)-F(a)}{x-a}=F^{\prime}(a)$
$\therefore F^{\prime}(x)=\frac{d}{d x}\left(\sqrt{9-x^{2}}\right)$
$=\frac{1(0-2 x)}{2 \sqrt{9-x^{2}}}=\frac{-x}{\sqrt{9-x^{2}}}$
$\quad F^{\prime}(1)=\frac{-1}{\sqrt{9-1}}=\frac{-1}{\sqrt{8}}=\frac{-1}{2 \sqrt{2}}$
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