Question

graph the following function and then find the specified limit. when necessary, state that the limit does not exist.
$g(x) = \

$$\begin{cases} x^2 & \\text{for } x \ eq -2 \\\\ x + 6 & \\text{for } x = -2 \\end{cases}$$

$ find $\lim\limits_{x \to -2} g(x)$

choose the correct graph below

find $\lim\limits_{x \to -2} g(x)$ select the correct choice below and fill in any answer boxes in your choice
\bigcirc a. $\lim\limits_{x \to -2} g(x) = \boxed{}$ (type an integer or a simplified fraction.)
\bigcirc b. the limit does not exist

Explanation:

Step1: Calculate left-hand limit

For \(x \to -2^-\), use \(G(x) = x^2\).
\(\lim_{x \to -2^-} x^2 = (-2)^2 = 4\)

Step2: Calculate right-hand limit

For \(x \to -2^+\), use \(G(x) = x + 6\).
\(\lim_{x \to -2^+} (x + 6) = -2 + 6 = 4\)

Step3: Check limit existence

Left-hand limit = Right-hand limit = 4. So limit exists.

Answer:

A. \(\lim_{x \to -2} G(x) = 4\)