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(C) For $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$,we have $f(x) = \sqrt{n}$. As $x \rightarrow 0$,$n \rightarrow \infty$,so $f(x) \approx \frac{

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is equal to $2$

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(C) For $x \in \left[\frac{1}{n+1}, \frac{1}{n}ight)$,we have $f(x) = \sqrt{n}$. As $x ightarrow 0$,$n ightarrow \infty$,so $f(x) \approx \frac{1}{\sqrt{x}}$.Given $\int_{x^2}^x \sqrt{\frac{1-t}{t}} dt Let $I(x) = \int_{x^2}^x \sqrt{\frac{1-t}{t}} dt$. Using the substitution $t = \sin^2 	heta$,$dt = 2\sin 	heta \cos 	heta d	heta$,we get $\int \sqrt{\frac{1-\sin^2 	heta}{\sin^2 	heta}} 2\sin 	heta \cos 	heta d	heta = \int 2\cos^2 	heta d	heta = \int (1 + \cos 2	heta) d	heta = 	heta + \frac{1}{2}\sin 2	heta = \arcsin \sqrt{t} + \sqrt{t(1-t)}$.Thus,$I(x) = [\arcsin \sqrt{t} + \sqrt{t(1-t)}]_{x^2}^x = \arcsin \sqrt{x} + \sqrt{x(1-x)} - \arcsin x - x\sqrt{1-x^2}$.As $x ightarrow 0$,$I(x) \approx \sqrt{x} + \sqrt{x} - 0 - 0 = 2\sqrt{x}$.Since $f(x) \approx \frac{1}{\sqrt{x}}$,we have $f(x)g(x) \approx \frac{1}{\sqrt{x}} \cdot (2\sqrt{x}) = 2$.By the Squeeze Theorem,$\lim_{x ightarrow 0} f(x)g(x) = 2$.

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