Since $x=0 \le 1$, $f(0) = 0$. The function is defined as $f(x) = x$ in the neighborhood of $x=0$. Therefore, the function is continuous at $x=0$.
We need to check the left-hand limit (LHL) and the right-hand limit (RHL) at $x=1$. LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$ RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$ Since LHL $\ne$ RHL, the function is not continuous at $x=1$.
Since $x=2 > 1$, $f(2) = 5$. The function is defined as $f(x) = 5$ in the neighborhood of $x=2$. Therefore, the function is continuous at $x=2$.
Since $x=5 > 1$, $f(5) = 5$. The function is defined as $f(x) = 5$ in the neighborhood of $x=5$. Therefore, the function is continuous at $x=5$.
The function is not continuous at $x=1$.
Final Answer: x=1
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