Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?
(A) The given function is $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$.
To check the continuity at $x=3$,we observe the definition of the function for $x > 1$.
For any $x > 1$,$f(x) = 2$.
At $x=3$,the value of the function is $f(3) = 2$.
The limit of the function as $x$ approaches $3$ is:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2) = 2$.
Since $\lim_{x \to 3} f(x) = f(3) = 2$,the function $f$ is continuous at $x=3$.
To check the continuity at $x=3$,we observe the definition of the function for $x > 1$.
For any $x > 1$,$f(x) = 2$.
At $x=3$,the value of the function is $f(3) = 2$.
The limit of the function as $x$ approaches $3$ is:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2) = 2$.
Since $\lim_{x \to 3} f(x) = f(3) = 2$,the function $f$ is continuous at $x=3$.
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