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To determine \(\lim\limits_{x \to -2} f(x)\), we need to analyze the piecewise function \( f(x) \). The function is defined differently depending on whether \( x \) is less than or greater than/equal to \(-2\):

\[
f(x) = 
\begin{cases} 
10 + 5x & \text{for } x < -2 \\
x^2 - 4 & \text{for } x \geq -2
\end{cases}
\]

### Step 1: Approach \( x = -2 \) from the left (left-hand limit)

When \( x \to -2^- \), we use the definition of \( f(x) \) for \( x < -2 \):
\[
f(x) = 10 + 5x
\]
Substituting \( x = -2 \):
\[
\lim_{x \to -2^-} f(x) = 10 + 5(-2) = 10 - 10 = 0
\]

### Step 2: Approach \( x = -2 \) from the right (right-hand limit)

When \( x \to -2^+ \), we use the definition of \( f(x) \) for \( x \geq -2 \):
\[
f(x) = x^2 - 4
\]
Substituting \( x = -2 \):
\[
\lim_{x \to -2^+} f(x) = (-2)^2 - 4 = 4 - 4 = 0
\]

### Step 3: Compare the left-hand and right-hand limits

Since both the left-hand limit and right-hand limit as \( x \to -2 \) are equal to 0, we conclude that the two-sided limit exists and is:

\[
\lim_{x \to -2} f(x) = 0
\]

### Conclusion:
The limit of \( f(x) \) as \( x \) approaches \(-2\) is \( 0 \).

---

Would you like more details or have any questions about this solution?  
Here are five related questions for further practice:

1. What is the limit of \( f(x) \) as \( x \to -2 \) if \( f(x) \) was defined differently for \( x \geq -2 \)?
2. How do you find the limit of a piecewise function at a point where the definition changes?
3. What is the value of \( f(-2) \) in this case? Is it continuous at \( x = -2 \)?
4. What is the limit of \( f(x) \) as \( x \to \infty \)?
5. How does the concept of one-sided limits help determine the existence of a two-sided limit?

**Tip:** Always remember to compare the left-hand and right-hand limits to determine whether a two-sided limit exists.

For the function f(x) shown below, determine lim(x → -2) f(x).

f(x) = { 10 + 5x for x < -2
 x^2 - 4 for x ≥ -2 }

Learn how to find the limit of a piecewise function as x approaches -2. This problem involves evaluating limits from both the left and right, using key concepts in calculus. Suitable for students in Grades 11-12.

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