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Let $f : R\rightarrow [0,\infty]$ be such that $\lim\limits_{x\to 5}f(x)$ exists and $\lim\limits_{x\to 5}\large\frac{(f(x))^2-9}{\sqrt{|x-5|}}$ = 0. Then $\lim\limits_{x\to 5}f(x)$ equals


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$\lim\limits_{x\to 5}\large\frac{(f(x))^2-9}{\sqrt{|x-5|}}$$=0$
$\lim\limits_{x\to 5}[(f(x))^2-9$$]=0$
$\Rightarrow \lim\limits_{x\to 5}(f)=3$
Hence (d) is the correct answer.
answered Dec 23, 2013 by sreemathi.v

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