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

$(a)\;1/24\qquad(b)\;1/5\qquad(c)\;-\sqrt{24}\qquad(d)\;None\;of\;these$

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$\lim\limits_{x\to 1}\large\frac{G(x)-G(1)}{x-1}=$$\lim\limits_{x\to 1}G'(x)=$$\lim\limits_{x\to 1}-\large\frac{1}{2\sqrt{25-x^2}}$$(-2x)$
$\Rightarrow \large\frac{1}{24}$
Hence (a) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
edited Mar 24, 2014 by sreemathi.v
 

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