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If $f(9)=9,f'(9)=4$ then $\lim\limits_{x\to 9}\large\frac{\sqrt{f(x)}-3}{\sqrt x-3}$ equals

$(a)\;5\qquad(b)\;4\qquad(c)\;3\qquad(d)\;2$

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Given that $f(9)=9,f'(9)=4$
Then,
$\lim\limits_{x\to 9}\large\frac{\sqrt{f(x)}-3}{\sqrt x-3}=$$\lim\limits_{x\to 9}\large\frac{(\sqrt{f(x)}-3)(\sqrt{f(x)}+3)}{(\sqrt x-3)(\sqrt x+3)}$ $\times \large\frac{\sqrt x+3}{\sqrt{f(x)}+3}$
$\Rightarrow \lim\limits_{x\to 9}\large\frac{f(x)-9}{x-9}\big[\large\frac{3+3}{3+3}\big]$
$\Rightarrow \lim\limits_{x\to 9}\large\frac{f(x)-f(9)}{x-9}.$$1$
$f'(9)=4$
Hence (b) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
edited Mar 18, 2014 by balaji.thirumalai
 

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