If G(x) = -sqrt{25 - x^2},then mathop{lim}lim | vedclass

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Question

(D) The expression $\mathop{\lim}\limits_{x 	o 1} \frac{G(x) - G(1)}{x - 1}$ represents the derivative of $G(x)$ at $x = 1$,denoted as $G'(1)$.Given

Vedclass · Accepted Answer

$\frac{1}{\sqrt{24}}$

Vedclass · Answer

(D) The expression $\mathop{\lim}\limits_{x 	o 1} \frac{G(x) - G(1)}{x - 1}$ represents the derivative of $G(x)$ at $x = 1$,denoted as $G'(1)$.Given $G(x) = -(25 - x^2)^{1/2}$.Differentiating with respect to $x$ using the chain rule:$G'(x) = -\frac{1}{2}(25 - x^2)^{-1/2} \cdot (-2x) = \frac{x}{\sqrt{25 - x^2}}$.Evaluating at $x = 1$:$G'(1) = \frac{1}{\sqrt{25 - 1^2}} = \frac{1}{\sqrt{24}}$.Thus,the limit is $\frac{1}{\sqrt{24}}$.

If $G(x) = -\sqrt{25 - x^2}$,then $\mathop{\lim}\limits_{x \to 1} \frac{G(x) - G(1)}{x - 1} = $

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