Question

for the function g(x) shown below, compute the following limits or state that they do not exist.

g(x) = \

$$\begin{cases} 0 & \\text{if } x \\leq -5 \\\\ \\sqrt{25 - x^2} & \\text{if } -5 < x < 5 \\\\ 2x & \\text{if } x \\geq 5 \\end{cases}$$

a. $\lim\limits_{x \to -5^-} g(x)$ b. $\lim\limits_{x \to -5^+} g(x)$ c. $\lim\limits_{x \to -5} g(x)$ d. $\lim\limits_{x \to 5^-} g(x)$ e. $\lim\limits_{x \to 5^+} g(x)$ f. $\lim\limits_{x \to 5} g(x)$

a. find $\lim\limits_{x \to -5^-} g(x)$. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

$\bigcirc$ a. $\lim\limits_{x \to -5^-} g(x) = \square$
$\bigcirc$ b. the limit does not exist.

Explanation:

Step1: Determine the domain for \( x \to -5^- \)

For \( x \to -5^- \), \( x < -5 \), so we use the part of the piecewise function where \( x \leq -5 \), which is \( g(x) = 0 \).

Step2: Evaluate the limit

Since \( g(x) = 0 \) for all \( x \leq -5 \), as \( x \) approaches \( -5 \) from the left (\( x \to -5^- \)), the limit is the value of \( g(x) \) in that interval, which is 0.

Answer:

A. \( \lim\limits_{x \to -5^-} g(x) = 0 \)