Question

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

Answer 1

Clearly form of the limit is $\frac{0}{0}$
Thus applying L-Hospital's rule,
$\underset{x\to 1}{lim}\frac{G\left(x\right)-G\left(1\right)}{x-1}=G^{\prime }\left(1\right)=\frac{x}{\sqrt{25-x^2}}{|}_{x=1}=\frac{1}{\sqrt{24}}$