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The given problem is asking to find a value of the constant \( a \) such that the limit \( \lim_{x \to -1} g(x) \) exists, where \( g(x) \) is defined piecewise:

\[
g(x) = \begin{cases} 
x^2 - 8x & \text{if } x \leq -1, \\
ax^3 - 3 & \text{if } x > -1.
\end{cases}
\]

### 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?

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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.

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 }


Determine the constant 'a' for which the limit of a piecewise function g(x) exists as x approaches -1. Learn how to find the value of 'a' by equating the left-hand and right-hand limits of g(x) and ensuring continuity. Suitable for high school students in Grades 10-12.

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