Is the function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$ continuous at $x=0$? At $x=1$? At $x=2$?
(N/A) The given function is $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$.
At $x=0$:
The function is defined at $x=0$ and $f(0) = 0$.
The limit is $\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0$.
Since $\lim_{x \to 0} f(x) = f(0)$,the function is continuous at $x=0$.
At $x=1$:
The function is defined at $x=1$ and $f(1) = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$.
Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$,the limit does not exist at $x=1$.
Therefore,the function is not continuous at $x=1$.
At $x=2$:
The function is defined at $x=2$ and $f(2) = 5$.
The limit is $\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5$.
Since $\lim_{x \to 2} f(x) = f(2)$,the function is continuous at $x=2$.
At $x=0$:
The function is defined at $x=0$ and $f(0) = 0$.
The limit is $\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0$.
Since $\lim_{x \to 0} f(x) = f(0)$,the function is continuous at $x=0$.
At $x=1$:
The function is defined at $x=1$ and $f(1) = 1$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$.
Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$,the limit does not exist at $x=1$.
Therefore,the function is not continuous at $x=1$.
At $x=2$:
The function is defined at $x=2$ and $f(2) = 5$.
The limit is $\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5$.
Since $\lim_{x \to 2} f(x) = f(2)$,the function is continuous at $x=2$.
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