Question

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

• A. $\frac{1}{24}$
• B. $\frac{1}{5}$
• C. $-\sqrt{24}$
• D. $\frac{1}{2\sqrt{6}}$

Answer 1

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

$\underset{x\to 1}{lim}\frac{-\sqrt{25-x^2}-(-\sqrt{24})}{x-1}$
$=\underset{x\to 1}{lim}\frac{\sqrt{24}-\sqrt{25-x^2}}{x-1}\times \frac{\sqrt{24}+\sqrt{25-x^2}}{\sqrt{24}+\sqrt{25-x^2}}$
$\underset{x\to 1}{lim}\frac{x^2-1}{(x-1)[\sqrt{24}+\sqrt{25-x^2}]}=\frac{2}{2\sqrt{24}}=\frac{1}{2\sqrt{6}}$