Math Problem Statement
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.
Solution
Let's go through the problem step by step.
The function is defined piecewise as:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Limits
Continuity
Formulas
$\lim_{x \to c^-} f(x)$
$\lim_{x \to c^+} f(x)$
$\lim_{x \to c} f(x)$
Theorems
Limit Definition
One-Sided Limits
Suitable Grade Level
Grades 11-12
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