We are tasked with finding the right-hand limit (denoted $\lim_{x \to 6^+}$) of the function $f(x) = \sqrt{x^2 + 8x + 16}$. The limit expression is:

$\lim_{x \to 6^+} \sqrt{x^2 + 8x + 16}$

Step 1: Simplify the expression inside the square root

We begin by simplifying the quadratic expression inside the square root:

$x^2 + 8x + 16 = (x + 4)^2$

Thus, the given function becomes:

$f(x) = \sqrt{(x + 4)^2}$

Step 2: Apply the square root property

The square root of a square is the absolute value of the expression inside:

$f(x) = |x + 4|$

Step 3: Analyze the behavior as $x \to 6^+$

Now, we evaluate the limit:

$\lim_{x \to 6^+} |x + 4|$

For values of $x$ slightly greater than 6, $x + 4$ is positive, so $|x + 4| = x + 4$. Thus, we can directly substitute $x = 6$ into the expression:

$\lim_{x \to 6^+} (x + 4) = 6 + 4 = 10$

Conclusion:

The right-hand limit is:

$\lim_{x \to 6^+} \sqrt{x^2 + 8x + 16} = 10$

Would you like a further breakdown of the steps, or do you have any specific questions?

Related Questions:

1. What is the difference between left-hand and right-hand limits?
2. How do absolute value functions affect limit calculations?
3. What are some typical behaviors of limits involving square roots?
4. How do we evaluate limits when the expression inside the square root is more complicated?
5. Can you describe cases when limits do not exist?

Tip:

When dealing with limits involving square roots, always check if the expression inside can be factored or simplified, as this can make finding the limit much easier.