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We are tasked with finding the limit \(\lim_{x \to 3} f(x)\) for the given piecewise function:

\[
f(x) = \begin{cases}
2x + 1 & \text{if } x < 3 \\
7 & \text{if } x = 3 \\
x + 4 & \text{if } x > 3
\end{cases}
\]

To evaluate this limit, we must check the left-hand limit \(\lim_{x \to 3^-} f(x)\) and the right-hand limit \(\lim_{x \to 3^+} f(x)\).

1. **Left-hand limit** (\(x \to 3^-\)):
   - For \(x < 3\), \(f(x) = 2x + 1\).
   - Substituting \(x = 3\) into this expression:
   \[
   \lim_{x \to 3^-} f(x) = 2(3) + 1 = 6 + 1 = 7.
   \]

2. **Right-hand limit** (\(x \to 3^+\)):
   - For \(x > 3\), \(f(x) = x + 4\).
   - Substituting \(x = 3\) into this expression:
   \[
   \lim_{x \to 3^+} f(x) = 3 + 4 = 7.
   \]

Since both the left-hand limit and right-hand limit are equal to 7, we conclude that:
\[
\lim_{x \to 3} f(x) = 7.
\]

Thus, the correct answer is **A) 7**.

Would you like more details, or do you have any questions?

Here are five related questions:
1. How would the limit change if the right-hand limit expression was different?
2. What if the piecewise function had a discontinuity at \(x = 3\)?
3. How would you define continuity based on this example?
4. What happens if you change the middle value \(f(3)\) from 7 to something else?
5. Can you explain how the piecewise function affects limits in general?

**Tip:** Always check both the left-hand and right-hand limits when dealing with piecewise functions to ensure they match for the limit to exist.

What is the limit \(\lim_{x \to 3} f(x)\) for the given piecewise function: \( f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ 7 & \text{if } x = 3 \\ x + 4 & \text{if } x > 3 \end{cases} \)?

Explore how to find the limit \(\lim_{x \to 3} f(x)\) for a piecewise function. This solution explains the evaluation of left-hand and right-hand limits for different intervals of a piecewise function. Perfect for students in Grades 10-12.

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