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!