If $G\left(x\right)=-\sqrt{25-x}$ then $\underset{x\to 1}{lim} \frac{G\left(x\right)-G\left(1\right)}{x-1}$ has the value 

Solution:

We have,

$\underset{x\to 1}{lim} \frac{G\left(x\right)-G\left(1\right)}{x-1}=G^{\mathrm{\prime }}\left(1\right)$    [By def. of derivative]

Now,

$G\left(x\right)=-\sqrt{25-x^2}\Rightarrow G^{\mathrm{\prime }}\left(x\right)=\frac{x}{\sqrt{25-x^2}}\Rightarrow G^{\mathrm{\prime }}\left(1\right)=\frac{1}{\sqrt{24}}$