Question

If $f\left(9\right)=0,f^{\prime }\left(9\right)=4$, then $\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}$ $=$

• A. 9
• B. 4
• C. 36
• D. None of these

Answer 1

The correct option is B 4

If $f\left(9\right)=9$ then solution will be ,

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\underset{x\to 9}{lim}\left(\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}\right)\left(\frac{\sqrt{f\left(x\right)}+3}{\sqrt{f\left(x\right)}+3}\right)\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\underset{x\to 9}{lim}\left(\frac{\sqrt{x}+3}{\sqrt{f\left(x\right)}+3}\right)\underset{x\to 9}{lim}\left(\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}\right)\left(\frac{\sqrt{f\left(x\right)}+3}{\sqrt{x}+3}\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\underset{x\to 9}{lim}\left(\frac{\sqrt{x}+3}{\sqrt{f\left(x\right)}+3}\right)\underset{x\to 9}{lim}\left(\frac{f\left(x\right)-9}{x-9}\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\underset{x\to 9}{lim}\left(\frac{\sqrt{x}+3}{\sqrt{f\left(x\right)}+3}\right)\underset{x\to 9}{lim}\left(\frac{f\left(x\right)-f\left(9\right)}{x-9}\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\frac{6}{6}\underset{x\to 9}{lim}\left(\frac{f\left(x\right)-f\left(9\right)}{x-9}\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\frac{6}{6}\underset{x\to 9}{lim}\left(f^{\prime }\right(9\left)\right)$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=1\times 4$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=4$