Discuss the continuity of the function $f,$ where $f$ is defined by $f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}$ at $x=3.$
(A) The given function is $f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}$.
To check the continuity at $x=3$,we evaluate the limit of the function as $x \to 3$ and compare it with the value of the function at $x=3$.
Since $3 > 1$,the function is defined by $f(x) = 4x$ in the neighborhood of $x=3$.
$1$. Value of the function at $x=3$:
$f(3) = 4(3) = 12$.
$2$. Limit of the function as $x \to 3$:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (4x) = 4(3) = 12$.
Since $\lim_{x \to 3} f(x) = f(3) = 12$,the function $f$ is continuous at $x=3$.
To check the continuity at $x=3$,we evaluate the limit of the function as $x \to 3$ and compare it with the value of the function at $x=3$.
Since $3 > 1$,the function is defined by $f(x) = 4x$ in the neighborhood of $x=3$.
$1$. Value of the function at $x=3$:
$f(3) = 4(3) = 12$.
$2$. Limit of the function as $x \to 3$:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (4x) = 4(3) = 12$.
Since $\lim_{x \to 3} f(x) = f(3) = 12$,the function $f$ is continuous at $x=3$.
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