Math Problem Statement
Using the given function below, evaluate each of the of the following limits, if it exists. Let g(x) = {
x if x < 0 x^2 if 0 < x ≤ 2 8 − x if x > 2
Evaluate each of the of the following limits, if it exists.
a. lim g(x) x→ 0^+ b. lim g(x) x→ 0^− c. lim g(x) x→ 0 d. lim g(x) x→ 2^− e. lim g(x) x→ 2
Solution
Given the piecewise function :
\begin{cases} x & \text{if } x < 0 \\ x^2 & \text{if } 0 < x \leq 2 \\ 8 - x & \text{if } x > 2 \end{cases}$$ We are tasked with evaluating various limits. Let’s go step by step. ### a. $$\lim_{x \to 0^+} g(x)$$ This represents the right-hand limit as $$x$$ approaches 0 from values greater than 0. When $$x > 0$$, the function definition is $$g(x) = x^2$$. Therefore, we use the $$x^2$$ expression to evaluate the limit: $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0$$ ### b. $$\lim_{x \to 0^-} g(x)$$ This represents the left-hand limit as $$x$$ approaches 0 from values less than 0. When $$x < 0$$, the function definition is $$g(x) = x$$. Therefore, we use the $$x$$ expression to evaluate the limit: $$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} x = 0$$ ### c. $$\lim_{x \to 0} g(x)$$ The limit exists if and only if both the left-hand and right-hand limits are equal. From parts (a) and (b), we found that: $$\lim_{x \to 0^+} g(x) = 0 \quad \text{and} \quad \lim_{x \to 0^-} g(x) = 0$$ Thus: $$\lim_{x \to 0} g(x) = 0$$ ### d. $$\lim_{x \to 2^-} g(x)$$ This represents the left-hand limit as $$x$$ approaches 2 from values less than 2. When $$0 < x \leq 2$$, the function definition is $$g(x) = x^2$$. Therefore, we evaluate the limit using the $$x^2$$ expression: $$\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4$$ ### e. $$\lim_{x \to 2} g(x)$$ The limit exists if both the left-hand and right-hand limits at $$x = 2$$ are equal. We already found: - $$\lim_{x \to 2^-} g(x) = 4$$ For the right-hand limit, when $$x > 2$$, the function is $$g(x) = 8 - x$$. Therefore: $$\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (8 - x) = 8 - 2 = 6$$ Since the left-hand limit ($$4$$) does not equal the right-hand limit ($$6$$), the overall limit does **not exist** at $$x = 2$$. --- ### Summary of Results: a. $$\lim_{x \to 0^+} g(x) = 0$$ b. $$\lim_{x \to 0^-} g(x) = 0$$ c. $$\lim_{x \to 0} g(x) = 0$$ d. $$\lim_{x \to 2^-} g(x) = 4$$ e. $$\lim_{x \to 2} g(x)$$ does not exist. --- Would you like any more details or have any questions? Here are 5 related questions for deeper exploration: 1. What is the value of $$\lim_{x \to 3} g(x)$$? 2. How does the function behave as $$x$$ approaches negative infinity? 3. For what values of $$x$$ is the function $$g(x)$$ continuous? 4. Can you determine any points where $$g(x)$$ is discontinuous? 5. How would the function $$g(x)$$ change if the middle piece were defined as $$x^3$$ instead of $$x^2$$? **Tip**: Always check both left-hand and right-hand limits when determining if a limit exists at a given point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Limits
Continuity
Formulas
Limit definition
Evaluating one-sided limits
Theorems
Existence of limits
Continuity at a point
Suitable Grade Level
Grade 11-12 (Advanced Pre-Calculus/Calculus)