Question

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

• A. $-\frac{1}{2\sqrt{6}}$
• B. $\frac{1}{5}$
• C. $-\frac{1}{\sqrt{6}}$
• D. $\frac{1}{\sqrt{6}}$

Answer 1

The correct option is A $-\frac{1}{2\sqrt{6}}$

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

$=\underset{x⟶1}{lim}\frac{-2x}{2\sqrt{25-x^2}}$ .... Applying L'Hospital's rule

$=\frac{-2}{2\sqrt{25-1}}=\frac{-1}{\sqrt{24}}=\frac{-1}{2\sqrt{6}}$