You are looking for the limit of the expression:

$\lim_{x \to 2} \frac{x^2 + 4x - 12}{x^2 - 2x}$

Step 1: Substitute $x = 2$ directly

Let's first try substituting $x = 2$ directly into the expression:

$\text{Numerator:} \quad 2^2 + 4(2) - 12 = 4 + 8 - 12 = 0$ $\text{Denominator:} \quad 2^2 - 2(2) = 4 - 4 = 0$

So, we get a $\frac{0}{0}$ indeterminate form. This means we need to simplify the expression further using algebraic methods.

Step 2: Factor both the numerator and denominator

Factor the numerator:

The numerator is $x^2 + 4x - 12$. We factor it as:

$x^2 + 4x - 12 = (x + 6)(x - 2)$

Factor the denominator:

The denominator is $x^2 - 2x$, which factors as:

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

Step 3: Simplify the expression

Now, substitute the factored forms back into the original expression:

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

We can cancel the common factor of $(x - 2)$ from the numerator and denominator (note that $x \neq 2$ to avoid division by zero):

$\frac{x + 6}{x}$

Step 4: Take the limit as $x \to 2$

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

$\frac{2 + 6}{2} = \frac{8}{2} = 4$

Final Answer:

$\lim_{x \to 2} \frac{x^2 + 4x - 12}{x^2 - 2x} = 4$

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Would you like more details on any of the steps or have any further questions?

Here are 5 related questions to explore:

1. How do you evaluate limits that result in indeterminate forms like $\frac{0}{0}$?
2. What is the importance of factoring when solving limits?
3. Can this method be used for higher degree polynomials? If so, how?
4. What other algebraic techniques can be used to solve limits besides factoring?
5. How do you determine when to apply L'Hôpital's Rule instead of factoring?

Tip: Whenever you encounter an indeterminate form like $\frac{0}{0}$, try factoring or simplifying the expression before attempting direct substitution.