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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Limits and Derivatives

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

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

1 Answer

$\lim\limits_{x\to 1}\large\frac{-\sqrt{25-x^2}-(-\sqrt{24})}{x-1}$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{\sqrt{24}-\sqrt{25-x^2}}{x-1}\times \frac{\sqrt{24}+\sqrt{25-x^2}}{\sqrt{24}+\sqrt{25-x^2}}$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{x^2-1}{(x-1)(\sqrt{24}+\sqrt{25}-x^2)}$
$\Rightarrow \large\frac{2}{2\sqrt{24}}$
$\Rightarrow \large\frac{1}{2\sqrt 6}$
Hence (d) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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