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Mathematics Question on limits and derivatives

If G(x)=−25−x2G(x) = - \sqrt{25 - x^2} then lim⁡x→1G(x)−G(1)x−1\displaystyle \lim_{x \to 1} \frac{G(x) - G(1)}{x - 1} has the value

A

124\frac{1}{24}

B

15\frac{1}{5}

C

−24 - \sqrt{24}

D

none of these

Answer

none of these

Explanation

Solution

lim⁡x→1−25−x2−(−24)x−1\displaystyle\lim_{x \to1} \frac{- \sqrt{25-x^{2}} - \left(-\sqrt{24}\right)}{x-1}
=lim⁡x→124−25−x2x−1×24+25−x224+25−x2= \displaystyle\lim_{x \to1} \frac{\sqrt{24} - \sqrt{25-x^{2}}}{x-1} \times\frac{\sqrt{24} + \sqrt{25-x^{2}}}{\sqrt{24} + \sqrt{25-x^{2}}}
=lim⁡x→1x2−1(x−1)[24+25−x2]= \displaystyle\lim_{x \to1} \frac{x^{2} - 1}{\left(x-1\right) \left[\sqrt{24} + \sqrt{25 -x^{2}}\right]}
=lim⁡x→1x+1[24+25−x2]= \displaystyle\lim_{x \to1} \frac{x+1}{ \left[\sqrt{24} + \sqrt{25-x^{2}}\right]}
=2224=126= \frac{2}{2\sqrt{24}} = \frac{1}{2\sqrt{6}}