Question

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

• A. $\frac{1}{24}$
• B. $\frac{1}{5}$
• C. $-\sqrt{24}$
• D. none of these

Answer 1

Answer: (D)

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