The given limit expression is:

$\lim_{{x \to 2}} \frac{{x^3 - 8}}{{3x^2 - 6x}}$

To solve this, we can follow these steps:

Step 1: Check for direct substitution

First, try substituting $x = 2$ into the expression:

$\frac{{2^3 - 8}}{{3(2^2) - 6(2)}} = \frac{{8 - 8}}{{3(4) - 12}} = \frac{0}{0}$

Since this results in an indeterminate form $\frac{0}{0}$, we need to simplify the expression further.

Step 2: Factor the numerator and denominator

• The numerator $x^3 - 8$ is a difference of cubes. This can be factored as:

$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

• The denominator $3x^2 - 6x$ can be factored as:

$3x^2 - 6x = 3x(x - 2)$

Step 3: Simplify the expression

Now substitute the factored forms into the original limit:

$\lim_{{x \to 2}} \frac{{(x - 2)(x^2 + 2x + 4)}}{{3x(x - 2)}}$

We can cancel the $(x - 2)$ terms from the numerator and denominator, assuming $x \neq 2$:

$\lim_{{x \to 2}} \frac{{x^2 + 2x + 4}}{{3x}}$

Step 4: Apply direct substitution again

Now, substitute $x = 2$ into the simplified expression:

$\frac{{2^2 + 2(2) + 4}}{{3(2)}} = \frac{{4 + 4 + 4}}{{6}} = \frac{{12}}{{6}} = 2$

Final Answer:

$\lim_{{x \to 2}} \frac{{x^3 - 8}}{{3x^2 - 6x}} = 2$

Would you like more details or further clarification?

Here are some additional questions related to this concept:

1. How do you solve limits involving indeterminate forms other than $\frac{0}{0}$?
2. What is the factorization process for cubes and squares in polynomials?
3. Can you explain why cancelling terms like $(x - 2)$ is valid?
4. How do you apply L'Hopital's Rule to solve limits?
5. What are common indeterminate forms in limits, and how can we simplify them?

Tip: Always check for factorization possibilities when you encounter indeterminate forms like $\frac{0}{0}$.