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Let $$f : [1,3] \to R$$ be a function satisfying $$\frac{x}{[x]} \le f(x) \le \sqrt{6-x}$$, for all $$x \neq 2$$ and $$f(2) = 1$$, where $$R$$ is the set of all real numbers and $$[x]$$ denotes the largest integer less than or equal to $$x$$. Statement 1: $$\lim_{x\to 2} f(x)$$ exists. Statement 2: $$f$$ is continuous at $$x = 2$$.
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