The given expression is:

$\lim_{x \to \infty} \sqrt{6x^2 - 2x + 1} - 4x + 2$

To solve this limit as $x \to \infty$, we'll follow these steps:

Step 1: Approximate the dominant term

When $x \to \infty$, the term $6x^2$ inside the square root dominates over $-2x + 1$. Thus, we approximate:

$\sqrt{6x^2 - 2x + 1} \approx \sqrt{6x^2} = \sqrt{6} \cdot x$

Step 2: Simplify the expression

Using the approximation, the limit becomes:

$\lim_{x \to \infty} \left( \sqrt{6} \cdot x - 4x + 2 \right)$

Step 3: Combine terms

Now, combine the $x$-terms:

$\lim_{x \to \infty} \left( (\sqrt{6} - 4) x + 2 \right)$

Step 4: Evaluate the limit

As $x \to \infty$, if $\sqrt{6} - 4$ is negative, the linear term will dominate and tend to negative infinity. Let's compute $\sqrt{6}$:

$\sqrt{6} \approx 2.449$

Thus, $\sqrt{6} - 4 \approx 2.449 - 4 = -1.551$.

Therefore, the limit is:

$\lim_{x \to \infty} -1.551x + 2 = -\infty$

Conclusion:

$\lim_{x \to \infty} \sqrt{6x^2 - 2x + 1} - 4x + 2 = -\infty$

Would you like more details or have any questions about this solution? Here are some related questions:

1. What are the steps for solving limits involving square roots?
2. How does the dominant term influence limits at infinity?
3. How can you use approximations to simplify complex limits?
4. What other types of limits behave similarly when approaching infinity?
5. How does this limit change if the constant term inside the square root changes?

Tip: Always look for the highest degree terms in polynomials when calculating limits as $x \to \infty$.