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(D) Given $\frac{x}{[x]} \le f(x) \le \sqrt{6 - x}$ for $x 
e 2$.For $x 	o 2^-$,$[x] = 1$. Thus,$\lim_{x 	o 2^-} \frac{x}{[x]} = \frac{2}{1} = 2$ a

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Statement $1$ is true,Statement $2$ is false.

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(D) Given $\frac{x}{[x]} \le f(x) \le \sqrt{6 - x}$ for $x 
e 2$.For $x 	o 2^-$,$[x] = 1$. Thus,$\lim_{x 	o 2^-} \frac{x}{[x]} = \frac{2}{1} = 2$ and $\lim_{x 	o 2^-} \sqrt{6 - x} = \sqrt{6 - 2} = 2$.By the Sandwich Theorem,$\lim_{x 	o 2^-} f(x) = 2$. So,Statement $1$ is true.For $x 	o 2^+$,$[x] = 2$. Thus,$\lim_{x 	o 2^+} \frac{x}{[x]} = \frac{2}{2} = 1$ and $\lim_{x 	o 2^+} \sqrt{6 - x} = 2$.By the Sandwich Theorem,$1 \le \lim_{x 	o 2^+} f(x) \le 2$. Since the left-hand limit is $2$ and the right-hand limit is not necessarily $2$,the limit $\lim_{x 	o 2} f(x)$ does not exist.Since $\lim_{x 	o 2} f(x)$ does not exist,$f$ is not continuous at $x = 2$. So,Statement $2$ is false.

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