You provided the limit expression:

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

To solve this limit, we can try direct substitution of $x = 2$ first. If it leads to an indeterminate form like $\frac{0}{0}$, we may need to apply algebraic techniques such as factoring or simplifying the expression.

Step 1: Direct Substitution

Substitute $x = 2$ into the expression:

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

This results in the indeterminate form $\frac{0}{0}$, so we must simplify the expression.

Step 2: Factor the Numerator

We will factor the quadratic expression $3x^2 - 4x - 4$.

1. Multiply the coefficient of $x^2$ (which is 3) by the constant term (which is -4), giving $3 \times (-4) = -12$.
2. Now, find two numbers that multiply to give -12 and add to give -4 (the coefficient of $x$). These numbers are -6 and 2.

So, we rewrite the quadratic as: $3x^2 - 6x + 2x - 4 = 0$ Now, factor by grouping: $3x(x - 2) + 2(x - 2) = (3x + 2)(x - 2)$

Step 3: Simplify the Expression

The original expression is now: $\frac{(3x + 2)(x - 2)}{x - 2}$ We can cancel out the $(x - 2)$ terms, leaving: $3x + 2$

Step 4: Evaluate the Limit

Now, substitute $x = 2$ into $3x + 2$: $3(2) + 2 = 6 + 2 = 8$

Thus, the limit is:

$\boxed{8}$

Would you like more details on any part of the process?

Here are 5 questions for further exploration:

1. How do we apply L'Hopital's Rule for limits with indeterminate forms?
2. What is the difference between factoring and simplifying in limit problems?
3. How do you find limits approaching infinity or negative infinity?
4. Can limits be calculated if the expression involves square roots?
5. What is the significance of one-sided limits like the left-hand limit here?

Tip: Always check if direct substitution works first before attempting algebraic manipulations in limit problems.