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If $f(9)=9$ and $f^{\prime}(9)=4$ then what is $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ equal to ?
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The correct answer is:
4
$\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{\mathrm{f}(\mathrm{x})}-3}{\sqrt{\mathrm{x}}-3}=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{1}{2 \sqrt{\mathrm{f}(\mathrm{x})}} \mathrm{f}^{\prime}(\mathrm{x})}{\frac{1}{2 \sqrt{\mathrm{x}}} .1}$
(ByL' Hospital rule)
$=\lim _{x \rightarrow 0} \frac{\mathrm{f}^{\prime}(\mathrm{x}) \times \sqrt{\mathrm{x}}}{\sqrt{\mathrm{f}(\mathrm{x})}}=\frac{\mathrm{f}^{\prime}(9) \times \sqrt{9}}{\sqrt{\mathrm{f}(9)}}$
$=\frac{4 \times 3}{\sqrt{9}}=\frac{4 \times 3}{3}=4$
(ByL' Hospital rule)
$=\lim _{x \rightarrow 0} \frac{\mathrm{f}^{\prime}(\mathrm{x}) \times \sqrt{\mathrm{x}}}{\sqrt{\mathrm{f}(\mathrm{x})}}=\frac{\mathrm{f}^{\prime}(9) \times \sqrt{9}}{\sqrt{\mathrm{f}(9)}}$
$=\frac{4 \times 3}{\sqrt{9}}=\frac{4 \times 3}{3}=4$
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