If f: R to [0, infty) is such that lim_{x to | vedclass

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(D) Given that $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$. Let $L = \lim_{x 	o 5} f(x)$. Since the limit exists,we can write $\lim_{x \

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$3$

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(D) Given that $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$. Let $L = \lim_{x 	o 5} f(x)$. Since the limit exists,we can write $\lim_{x 	o 5} ((f(x))^2 - 9) = L^2 - 9$. If $L^2 - 9 
eq 0$,then the limit $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}}$ would be of the form $\frac{k}{0}$ (where $k 
eq 0$),which would be $\infty$. Since the given limit is $0$,the numerator must approach $0$ as $x 	o 5$. Therefore,$L^2 - 9 = 0$,which implies $L^2 = 9$. Since the codomain of $f$ is $[0, \infty)$,$L$ must be non-negative. Thus,$L = 3$.

If $f: R \to [0, \infty)$ is such that $\lim_{x \to 5} f(x)$ exists and $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$,then $\lim_{x \to 5} f(x)$ equals:

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