Question

If $G\left(x\right)=-\sqrt{25-x^2}$, then $\underset{x\to 1}{lim}\frac{G\left(x\right)-G\left(1\right)}{x-1}$ is

• A. $\frac{1}{24}$
• B. $\frac{1}{5}$
• C. $-\sqrt{24}$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$\underset{x\to 1}{lim}\frac{G\left(x\right)-G\left(1\right)}{x-1}$, $\frac{0}{0}$ form

$=\underset{x\to 1}{lim}\frac{G^{\prime }\left(x\right)-0}{1-0}$, Using L-hospital rule

$=G^{\prime }\left(1\right)$

Now $G^{\prime }\left(x\right)=-\frac{1}{2\sqrt{25-x^2}}(-2x)$, differentiate using chain rule

$\Rightarrow G^{\prime }\left(x\right)=\frac{x}{\sqrt{25-x^2}}$

$\therefore G^{\prime }\left(1\right)=\frac{1}{\sqrt{24}}$, which is required limit