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. $\lim\limits_{x \to 5} g(x)$

c. 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
a. $\lim\limits_{x \to -5} g(x) = 0$
b. the limit does not exist.

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

Explanation:

Part d: $\boldsymbol{\lim_{x \to 5^-} g(x)}$

Step 1: Identify the relevant piece

For $x \to 5^-$, we consider values of $x$ just less than 5. The piece of $g(x)$ for $-5 < x < 5$ is $g(x) = \sqrt{25 - x^2}$.

Step 2: Substitute $x = 5$ into the piece

Evaluate $\sqrt{25 - (5)^2}$. Simplify inside the square root: $25 - 25 = 0$. So $\sqrt{0} = 0$. Wait, no—wait, when $x$ approaches $5$ from the left, we use the middle piece. Wait, $x \to 5^-$ means $x$ is less than 5, so the middle piece $-5 < x < 5$? Wait, no, the middle piece is $-5 < x < 5$, and the right piece is $x \geq 5$. Wait, $x \to 5^-$ is $x$ approaching 5 from values less than 5, so we use the middle piece: $g(x) = \sqrt{25 - x^2}$ for $-5 < x < 5$. So substitute $x = 5$ into this piece (since the limit as $x$ approaches 5 from the left, we use the piece defined for $x$ near 5 but less than 5). So $\lim_{x \to 5^-} \sqrt{25 - x^2} = \sqrt{25 - 5^2} = \sqrt{25 - 25} = \sqrt{0} = 0$? Wait, no, wait: $25 - x^2$ when $x = 5$ is 0, so the square root is 0. Wait, but let's check again. The middle piece is $-5 < x < 5$, so as $x$ approaches 5 from the left (values like 4.9, 4.99, etc.), we use $g(x) = \sqrt{25 - x^2}$. So plugging in $x = 5$ (even though $x = 5$ is not in the middle piece, the limit is about the behavior near $x = 5$ from the left). So $\lim_{x \to 5^-} \sqrt{25 - x^2} = \sqrt{25 - 5^2} = 0$. Wait, but let's compute it properly. Let $x$ approach 5 from the left, so $x = 5 - h$ where $h$ is a small positive number approaching 0. Then $25 - x^2 = 25 - (5 - h)^2 = 25 - (25 - 10h + h^2) = 10h - h^2$. As $h \to 0$, $10h - h^2 \to 0$, so $\sqrt{10h - h^2} \to 0$. So the limit is 0.

Answer:

$\lim_{x \to 5^-} g(x) = \boxed{0}$ (so option A with 0)