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Let's go through the problem step by step.

The function $ g(x) $ is defined piecewise as:
\[
g(x) = 
\begin{cases}
0 & \text{if } x \leq -4 \\
\sqrt{16 - x^2} & \text{if } -4 < x < 4 \\
2x & \text{if } x \geq 4
\end{cases}
\]

The limits being evaluated are:
1. $ \lim\limits_{x \to -4^-} g(x) $
2. $ \lim\limits_{x \to -4^+} g(x) $
3. $ \lim\limits_{x \to -4} g(x) $
4. $ \lim\limits_{x \to 4^-} g(x) $
5. $ \lim\limits_{x \to 4^+} g(x) $
6. $ \lim\limits_{x \to 4} g(x) $

Let's analyze these limits:

### 1. $ \lim\limits_{x \to -4^-} g(x) $
For $ x \leq -4 $, the function $ g(x) = 0 $. Therefore, $ \lim\limits_{x \to -4^-} g(x) = 0 $.

### 2. $ \lim\limits_{x \to -4^+} g(x) $
For $ -4 < x < 4 $, the function $ g(x) = \sqrt{16 - x^2} $. When approaching $ x = -4 $ from the right (i.e., $ x \to -4^+ $), we get:
\[
\lim\limits_{x \to -4^+} \sqrt{16 - x^2} = \sqrt{16 - (-4)^2} = \sqrt{16 - 16} = 0
\]
So, $ \lim\limits_{x \to -4^+} g(x) = 0 $.

### 3. $ \lim\limits_{x \to -4} g(x) $
Since both one-sided limits exist and are equal, the two-sided limit also exists:
\[
\lim\limits_{x \to -4} g(x) = 0
\]

### 4. $ \lim\limits_{x \to 4^-} g(x) $
For $ -4 < x < 4 $, the function is $ g(x) = \sqrt{16 - x^2} $. As $ x \to 4^- $:
\[
\lim\limits_{x \to 4^-} \sqrt{16 - x^2} = \sqrt{16 - 4^2} = \sqrt{16 - 16} = 0
\]
So, $ \lim\limits_{x \to 4^-} g(x) = 0 $.

### 5. $ \lim\limits_{x \to 4^+} g(x) $
For $ x \geq 4 $, the function is $ g(x) = 2x $. As $ x \to 4^+ $:
\[
\lim\limits_{x \to 4^+} 2x = 2(4) = 8
\]
So, $ \lim\limits_{x \to 4^+} g(x) = 8 $.

### 6. $ \lim\limits_{x \to 4} g(x) $
Since the left-hand and right-hand limits are not equal (0 and 8), the two-sided limit does not exist:
\[
\lim\limits_{x \to 4} g(x) \text{ does not exist}.
\]

#### Answer to the specific question:
You are asked to find $ \lim\limits_{x \to -4^-} g(x) $. As we found earlier, this limit equals $ 0 $, so the correct choice is $ \boxed{0} $.

Would you like more details on any specific step or have any questions?

Here are five related questions to deepen your understanding:
1. How do you evaluate limits for piecewise functions at boundary points?
2. What conditions must be satisfied for a two-sided limit to exist?
3. Why does the limit $ \lim\limits_{x \to 4} g(x) $ not exist in this case?
4. How do one-sided limits help in analyzing discontinuities in functions?
5. What is the difference between continuous and discontinuous functions in terms of limits?

**Tip:** When dealing with piecewise functions, always check the behavior of the function from both sides of a point to determine the existence of limits.

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

$$
 g(x) = 
\begin{cases} 
0 & \text{if } x \leq -4 \\
 \sqrt{16 - x^2} & \text{if } -4 < x < 4 \\
 2x & \text{if } x \geq 4
 \end{cases} 
$$

Compute the following limits: 
(a) $\lim_{x \to -4^-} g(x)$ 
(b) $\lim_{x \to -4^+} g(x)$ 
(c) $\lim_{x \to -4} g(x)$ 
(d) $\lim_{x \to 4^-} g(x)$ 
(e) $\lim_{x \to 4^+} g(x)$ 
(f) $\lim_{x \to 4} g(x)$. 

Find $\lim_{x \to -4^-} g(x)$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Learn how to compute the limits for the piecewise function $g(x)$, including one-sided and two-sided limits. This problem covers key calculus concepts, such as limit definitions and one-sided limits, suitable for students in Grades 11-12.

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