Math Problem Statement
Determine a value of the constant a for which lim x→−1 g(x) exists and state the value of the limit, if possible, where g(x) is defined as:
g(x) = { x^2 - 8x if x ≤ -1
a x^3 - 3 if x > -1 }
Solution
The given problem is asking to find a value of the constant such that the limit exists, where is defined piecewise:
x^2 - 8x & \text{if } x \leq -1, \\ ax^3 - 3 & \text{if } x > -1. \end{cases}$$ ### Steps: 1. **Left-hand limit** ($$x \to -1^-$$): For $$x \leq -1$$, the function is $$g(x) = x^2 - 8x$$. $$\lim_{x \to -1^-} g(x) = (-1)^2 - 8(-1) = 1 + 8 = 9.$$ 2. **Right-hand limit** ($$x \to -1^+$$): For $$x > -1$$, the function is $$g(x) = ax^3 - 3$$. $$\lim_{x \to -1^+} g(x) = a(-1)^3 - 3 = -a - 3.$$ 3. **Equating the two limits**: For the limit $$\lim_{x \to -1} g(x)$$ to exist, the left-hand limit and the right-hand limit must be equal: $$9 = -a - 3.$$ Solving for $$a$$: $$9 + 3 = -a \quad \Rightarrow \quad a = -12.$$ Thus, the value of $$a$$ is $$-12$$. ### Final Answer: - The limit exists for $$a = -12$$, and the value of the limit is $$9$$. Would you like more details or have any questions? --- Here are some related questions: 1. What happens if $$a$$ has a different value? 2. How would you compute the limit if the function was not piecewise? 3. Can the limit exist if the left-hand and right-hand limits are not equal? 4. What is the general condition for limits to exist for piecewise functions? 5. How can this problem be applied to real-world scenarios involving piecewise functions? **Tip:** When solving limits for piecewise functions, always ensure the left-hand and right-hand limits match at the point of interest.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Continuity
Formulas
lim x→c f(x)
Piecewise function continuity condition
Theorems
Limit existence condition
Continuity of piecewise functions
Suitable Grade Level
Grades 10-12
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