The given problem asks to evaluate the limit:

$\lim_{x \to 6} \frac{\sqrt{x} - 2}{x - 6}$

Step 1: Direct Substitution

Substitute $x = 6$ into the expression:

$\frac{\sqrt{6} - 2}{6 - 6} = \frac{\sqrt{6} - 2}{0}$

This results in an indeterminate form $\frac{0}{0}$. To resolve this, we can try to simplify the expression using algebraic techniques, such as rationalizing the numerator.

Step 2: Rationalizing the Numerator

Multiply both the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x} + 2$:

$\frac{\sqrt{x} - 2}{x - 6} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 6)(\sqrt{x} + 2)}$

Using the difference of squares in the numerator:

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

So, the expression becomes:

$\frac{x - 4}{(x - 6)(\sqrt{x} + 2)}$

Step 3: Factor and Cancel Terms

Now, we factor the numerator $x - 4$:

$x - 4 = (x - 6)(x + 2)$

Thus, the expression simplifies to:

$\frac{(x - 6)(x + 2)}{(x - 6)(\sqrt{x} + 2)}$

Cancel out the common factor of $x - 6$ from the numerator and denominator (for $x \neq 6$):

$\frac{x + 2}{\sqrt{x} + 2}$

Step 4: Apply the Limit

Now, substitute $x = 6$ into the simplified expression:

$\frac{6 + 2}{\sqrt{6} + 2} = \frac{8}{\sqrt{6} + 2}$

Thus, the exact answer is:

$\frac{8}{\sqrt{6} + 2}$

Would you like further details or have any questions?

Here are five related questions you might consider:

1. How do you know when to rationalize the numerator in a limit problem?
2. What are other methods for resolving the indeterminate form $\frac{0}{0}$?
3. How would you handle limits involving higher powers or cube roots?
4. Can you apply L'Hopital's Rule to this problem, and what would that look like?
5. How would you solve a limit involving trigonometric functions?

Tip: When encountering a limit that results in an indeterminate form, always consider algebraic manipulation techniques, such as factoring or rationalizing, before applying more advanced methods like L'Hopital's Rule.