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Question: If \(G(x) = - \sqrt{25 - x^{2}}\), then ( \lim_{x \rightarrow 1}\frac{G(x) - G(1)}{x - 1} ) equals...

If G(x)=−25−x2G(x) = - \sqrt{25 - x^{2}}, then ( \lim_{x \rightarrow 1}\frac{G(x) - G(1)}{x - 1} ) equals

A

1/24

B

1/5

C

−24- \sqrt{24}

D

None of these

Answer

None of these

Explanation

Solution

lim⁡x→1G(x)−G(1)x−1=lim⁡x→1−25−x2+24x−1\lim_{x \rightarrow 1}\frac{G(x) - G(1)}{x - 1} = \lim_{x \rightarrow 1}\frac{- \sqrt{25 - x^{2}} + \sqrt{24}}{x - 1}

[Multiply both numerator and denominator by

(24+25−x2\sqrt{24} + \sqrt{25 - x^{2}})]

=lim⁡x→1x+124+25−x2=124= \lim_{x \rightarrow 1}\frac{x + 1}{\sqrt{24} + \sqrt{25 - x^{2}}} = \frac{1}{\sqrt{24}}

Alternative method: By L'-Hospital rule,

lim⁡x→1G′(x)1=lim⁡x→1−1(−2x)225−x2=124\lim_{x \rightarrow 1}\frac{G^{'}(x)}{1} = \lim_{x \rightarrow 1}\frac{- 1( - 2x)}{2\sqrt{25 - x^{2}}} = \frac{1}{\sqrt{24}}