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To evaluate the limit using continuity, let's analyze the given expression:

\[
\lim_{x \to 5} x \sqrt{29 - x^2}
\]

### Step 1: Check for continuity
The function \( f(x) = x \sqrt{29 - x^2} \) is a combination of the functions \(g(x) = x\) and \(h(x) = \sqrt{29 - x^2}\). The square root function \(h(x) = \sqrt{29 - x^2}\) is continuous where its argument is non-negative, i.e., \(29 - x^2 \geq 0\). This implies \( -\sqrt{29} \leq x \leq \sqrt{29} \approx 5.385\). Since \(x = 5\) lies within this interval, the function \(f(x)\) is continuous at \(x = 5\).

### Step 2: Evaluate the limit using continuity
Since the function is continuous at \(x = 5\), we can directly substitute \(x = 5\) into the expression:
\[
f(5) = 5 \sqrt{29 - 5^2} = 5 \sqrt{29 - 25} = 5 \sqrt{4} = 5 \cdot 2 = 10
\]

### Conclusion
\[
\lim_{x \to 5} x \sqrt{29 - x^2} = 10
\]

Would you like more details or have any questions about this process?

### Related Questions
1. How would the solution change if the limit was taken as \(x \to -5\)?
2. What would be the limit if \(x \to 0\) for the same expression?
3. Is the function \(f(x) = x \sqrt{29 - x^2}\) defined for all real numbers \(x\)?
4. What happens to \(f(x)\) as \(x \to \sqrt{29}\)?
5. Can you evaluate the derivative of \(f(x) = x \sqrt{29 - x^2}\) at \(x = 5\)?

### Tip
When a function is continuous at a point, evaluating the limit is as simple as substituting the point into the function!

Use continuity to evaluate the limit lim(x→5) x√(29−x^2).

Evaluate the limit lim(x→5) x√(29−x^2) using continuity principles. This problem involves limits, continuity, and square root functions, making it suitable for Grades 11-12 students.

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