Question

If $f\left(9\right)= 9 , f'\left(9\right) = 4$ then $\lim_{x\to9} \frac{\sqrt{f\left(x\right)} -3}{\sqrt{x} - 3} =$

• A. 9
• B. 3
• C. 4
• D. 27

Answer 1

Answer: (C)

4

Answer 2

We have $\lim_{x\to9} \frac{\sqrt{f\left(x\right)} -3}{\sqrt{x} - 3} \left(\frac{0}{0} from\right)$
Applying L-Hospital's Rule, we have
$\lim _{x\to 9} \frac{\frac{1}{2} \left(f\left(x\right)\right)^{\frac{-1}{2}} f'\left(x\right)}{\frac{1}{2}\left(x\right)^{\frac{-1}{2}}} = \lim _{x\to 9} \frac{x^{\frac{1}{2}}f'\left(x\right)}{\left(f\left(x\right)\right)^{\frac{1}{2}}}$
$= \frac{\left(9\right)^{\frac{1}{2}} f'\left(9\right)}{\left(f\left(9\right)\right)^{\frac{1}{2}} } = \frac{3.\left(4\right)}{\left(9\right)^{\frac{1}{2}}} = 4$