Question

$f(x)=\

$$\begin{cases} 3x+4 & \\text{if } x<2 \\\\ (x-2)^2+3 & \\text{if } x\\geq2 \\end{cases}$$

$
determine if $f(x)$ is continuous at $x=2$. if not, select the option with the correct reasoning as to why not.

not continuous at $x=2$ because $\lim_{x \to 2} f(x)$ does not exist

not continuous at $x=2$ because the limit exists but is unequal to $f(2)$

continuous at $x=2$

not continuous at $x=2$ because $f(2)$ is undefined

Explanation:

Step1: Calculate left-hand limit

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x+4) = 3(2)+4 = 10$

Step2: Calculate right-hand limit

$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x-2)^2+3 = (2-2)^2+3 = 3$

Step3: Compare limits and $f(2)$

$f(2) = (2-2)^2+3 = 3$. Since $\lim_{x \to 2^-} f(x)
eq \lim_{x \to 2^+} f(x)$, $\lim_{x \to 2} f(x)$ does not exist.

Answer:

Not continuous at $x=2$ because $\lim_{x \to 2} f(x)$ does not exist