The problem is asking to find the value of the following limit:

$\lim_{{x \to 6}} \frac{x^2 - 36}{x - 6}$

Step-by-Step Solution:

1. Recognize the Indeterminate Form: When $x = 6$ is directly substituted into the expression, the numerator becomes: $x^2 - 36 = 6^2 - 36 = 36 - 36 = 0$ The denominator becomes: $x - 6 = 6 - 6 = 0$ So, the expression is of the indeterminate form $\frac{0}{0}$.

2. Factor the Numerator: The expression $x^2 - 36$ is a difference of squares, which can be factored as: $x^2 - 36 = (x - 6)(x + 6)$ Thus, the limit becomes: $\lim_{{x \to 6}} \frac{(x - 6)(x + 6)}{x - 6}$

3. Simplify the Expression: Cancel the common factor $x - 6$ from both the numerator and denominator: $\lim_{{x \to 6}} (x + 6)$

4. Substitute $x = 6$: After simplification, substitute $x = 6$ into the remaining expression: $6 + 6 = 12$

Thus, the value of the limit is $\boxed{12}$, which corresponds to option d.

Let me know if you would like any further details or clarifications!

Relative questions:

1. What is an indeterminate form in calculus?
2. How do you factor the difference of squares?
3. Why is it important to simplify expressions before evaluating limits?
4. What other techniques can be used to solve limits?
5. How do you apply L'Hopital's Rule in limit problems?

Tip:

Always check for common factorizations in limit problems, as this often helps to resolve indeterminate forms like $\frac{0}{0}$.