To evaluate the limit

$\lim_{{x \to \infty}} \frac{6x^2 + 2x + 1}{6x^2 - 3x + 1},$

we can analyze the behavior of the function as $x$ approaches infinity by dividing every term in both the numerator and denominator by $x^2$, the highest power of $x$ present in the expression.

So, we have:

$\lim_{{x \to \infty}} \frac{6x^2 + 2x + 1}{6x^2 - 3x + 1} = \lim_{{x \to \infty}} \frac{\frac{6x^2}{x^2} + \frac{2x}{x^2} + \frac{1}{x^2}}{\frac{6x^2}{x^2} - \frac{3x}{x^2} + \frac{1}{x^2}}.$

Simplifying each term by dividing:

$= \lim_{{x \to \infty}} \frac{6 + \frac{2}{x} + \frac{1}{x^2}}{6 - \frac{3}{x} + \frac{1}{x^2}}.$

As $x \to \infty$, the terms $\frac{2}{x}$, $\frac{1}{x^2}$, and $\frac{3}{x}$ all approach zero. This leaves:

$= \frac{6 + 0 + 0}{6 - 0 + 0} = \frac{6}{6} = 1.$

Therefore, the answer is:

$\lim_{{x \to \infty}} \frac{6x^2 + 2x + 1}{6x^2 - 3x + 1} = 1.$

Would you like further details on any step or have any questions?

Here are some related questions:

1. What happens if the highest degree terms in the numerator and denominator have different coefficients?
2. How would the limit change if the powers in the numerator and denominator were different?
3. Can L'Hôpital's Rule be applied to solve this problem as well?
4. How does this approach apply to other rational functions?
5. What if we want to find limits as $x \to -\infty$?

Tip: For rational functions where the degrees of the numerator and denominator are equal, the limit as $x \to \infty$ is the ratio of the leading coefficients.