We are asked to find the limit:

$\lim_{x \to 2} \frac{3x^2 - 8x + 4}{5x^2 - 14x + 8}$

Step 1: Direct Substitution

First, substitute $x = 2$ into both the numerator and the denominator to see if the limit can be evaluated directly.

For the numerator: $3(2)^2 - 8(2) + 4 = 3(4) - 16 + 4 = 12 - 16 + 4 = 0$

For the denominator: $5(2)^2 - 14(2) + 8 = 5(4) - 28 + 8 = 20 - 28 + 8 = 0$

Since both the numerator and denominator are zero when $x = 2$, we have an indeterminate form $\frac{0}{0}$, so we need to apply factoring or L'Hôpital's Rule.

Step 2: Factoring the expression

Let's factor both the numerator and the denominator.

The numerator $3x^2 - 8x + 4$ can be factored as: $3x^2 - 8x + 4 = (3x - 2)(x - 2)$

The denominator $5x^2 - 14x + 8$ can be factored as: $5x^2 - 14x + 8 = (5x - 4)(x - 2)$

Now the limit expression becomes: $\lim_{x \to 2} \frac{(3x - 2)(x - 2)}{(5x - 4)(x - 2)}$

Step 3: Canceling common terms

Since $(x - 2)$ is common in both the numerator and denominator, we cancel it out (note that $x \neq 2$, so this cancellation is valid): $\lim_{x \to 2} \frac{3x - 2}{5x - 4}$

Step 4: Substituting $x = 2$

Now substitute $x = 2$ into the simplified expression: $\frac{3(2) - 2}{5(2) - 4} = \frac{6 - 2}{10 - 4} = \frac{4}{6} = \frac{2}{3}$

Final Answer:

$\lim_{x \to 2} \frac{3x^2 - 8x + 4}{5x^2 - 14x + 8} = \frac{2}{3}$

Let me know if you'd like more details or have any questions.

Here are some related questions to practice:

1. What would happen if you tried to apply L'Hôpital's Rule instead of factoring?
2. How can you find the limit if the numerator and denominator were both linear polynomials?
3. What is the geometric interpretation of a limit that gives a fraction like $\frac{2}{3}$?
4. How can factoring help in limits when the indeterminate form $\frac{0}{0}$ occurs?
5. What is the limit of $\frac{x^2 - 4}{x - 2}$ as $x \to 2$?

Tip: Always check if a limit results in an indeterminate form before deciding on the method (factoring, L'Hôpital's Rule, etc.).