Question

Let $f\left(x\right)=\frac{1}{\sqrt{18-x^2}}$, then the value of $\underset{x\to 3}{lim}\frac{f\left(x\right)-f\left(3\right)}{x-3}$ is

• A. $0$
• B. $-\frac{1}{9}$
• C. $-\frac{1}{3}$
• D. none of these

Answer 1

Answer: (D)

none of these

Let, $L=\underset{x\to 3}{lim}\frac{f\left(x\right)-f\left(3\right)}{x-3},\frac{0}{0}$ form
Hence using L-hospital's rule,
$L=\underset{x\to 3}{lim}\frac{f^{\prime }\left(x\right)}{1}=f^{\prime }\left(3\right)$
Now given $f\left(x\right)=\frac{1}{\sqrt{18-x^2}}$
$\therefore f^{\prime }\left(x\right)=\frac{1}{18-x^2}.\frac{1}{\sqrt{18-x^2}}\left(x\right)$
$\Rightarrow f^{\prime }\left(3\right)=\frac{1}{9}$
Therefore, the required limit is, $L=\frac{1}{9}$
Hence, option 'D' is correct.