Question

If $f\left(x\right)=-\sqrt{25-x^2}$, then find ${lim}_{x\to 1}\left(\frac{f\left(x\right)-f\left(1\right)}{x-1}\right)$

Answer 1

Given:

$f\left(x\right)=-\sqrt{25-x^2}$

${lim}_{x\to 1}\left(\frac{f\left(x\right)-f\left(1\right)}{x-1}\right)$

$=\underset{x\to 1}{lim}\left(\frac{-\sqrt{25-x^2}+\sqrt{24}}{x-1}\right)$

Put $x=1$

$=\left(\frac{-\sqrt{25-1^2}+\sqrt{24}}{1-1}\right)$

$=\left(\frac{-\sqrt{24}+\sqrt{24}}{0}\right)$

$=\frac{0}{0}$

As it is of type $\left(\frac{0}{0}\right)$, applying L-hospital rule for

$=\underset{x\to 1}{lim}\left(\frac{-\sqrt{25-x^2}+\sqrt{24}}{x-1}\right)$

By differentiating the both numerator and the denomitors with respect to '$x$'

$=\underset{x\to 1}{lim}(\frac{-1}{2\sqrt{25-x^2}}\times \frac{(-2x)}{1})$ [$\because \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}},\frac{d}{dx}{x}^n=n{x}^{n-1}$]

$={lim}_{x\to 1}\frac{x}{\sqrt{25-x^2}}$

$=\frac{1}{\sqrt{25-1^2}}$

$=\frac{1}{\sqrt{24}}$

$=\frac{1}{2\sqrt{6}}$

$\therefore \text{If}f\left(x\right)=-\sqrt{25-x^2}\text{then}{lim}_{x\to 1}\left(\frac{f\left(x\right)-f\left(1\right)}{x-1}\right)=\frac{1}{2\sqrt{6}}$