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. (limlimits_{x \to -5^-} g(x))
b. (limlimits_{x \to -5^+} g(x))
c. (limlimits_{x \to -5} g(x))
d. (limlimits_{x \to 5^-} g(x))
e. (limlimits_{x \to 5^+} g(x))
f. (limlimits_{x \to 5} g(x))

a. (limlimits_{x \to -5^-} g(x) = 0)
b. the limit does not exist.

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

a. (limlimits_{x \to -5^+} g(x) = square)
b. the limit does not exist.

Explanation:

Part b:

Step1: Determine the relevant piece.

For \( x \to -5^+ \), we use the piece \( g(x)=\sqrt{25 - x^2} \) (since \( -5 < x < 5 \) includes values just above -5).

Step2: Substitute \( x = -5 \) into the piece.

Substitute \( x=-5 \) into \( \sqrt{25 - x^2} \): \( \sqrt{25 - (-5)^2}=\sqrt{25 - 25}=0 \).

Answer:

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