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If $f(x)=\large\frac{1}{\sqrt{18-x^2}}$ then $\lim\limits_{x\to 3}\large\frac{f(x)-f(3)}{x-3}$ is

$(a)\;0\qquad(b)\;-1/9\qquad(c)\;-1/3\qquad(d)\;None\;of\;these$

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$\lim\limits_{x\to 3}\large\frac{f(x)-f(3)}{x-3}$$=f'(3)$
Now,$f'(x)=\large\frac{-1/2(-2x)}{(18-x^2)^{3/2}}$
$\Rightarrow f'(3)=\large\frac{1}{9}$
Hence (d) is the correct option.
answered Dec 30, 2013 by sreemathi.v
 

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