We are asked to evaluate the following limit as $n \to \infty$:

$\lim_{n \to \infty} \left( \frac{6n^3 + n^2 - 6n + 5}{n^2 - 4n + 1} - 6n \right)$

Step-by-step Solution:

1. Combine the terms under a common denominator:

   We first combine the expression into a single fraction:

   $\frac{6n^3 + n^2 - 6n + 5}{n^2 - 4n + 1} - 6n = \frac{6n^3 + n^2 - 6n + 5 - 6n(n^2 - 4n + 1)}{n^2 - 4n + 1}$

2. Expand $6n(n^2 - 4n + 1)$:

   Expand the term involving $6n$:

   $6n(n^2 - 4n + 1) = 6n^3 - 24n^2 + 6n$

   Now substitute this into the expression:

   $\frac{6n^3 + n^2 - 6n + 5 - (6n^3 - 24n^2 + 6n)}{n^2 - 4n + 1}$

3. Simplify the numerator:

   Combine like terms in the numerator:

   $(6n^3 - 6n^3) + (n^2 + 24n^2) + (-6n - 6n) + 5 = 25n^2 - 12n + 5$

   So the expression becomes:

   $\frac{25n^2 - 12n + 5}{n^2 - 4n + 1}$

4. Evaluate the limit as $n \to \infty$:

   Now, divide both the numerator and denominator by $n^2$ (the highest power of $n$):

   $\frac{25n^2 - 12n + 5}{n^2 - 4n + 1} = \frac{25 - \frac{12}{n} + \frac{5}{n^2}}{1 - \frac{4}{n} + \frac{1}{n^2}}$

   As $n \to \infty$, the terms with $\frac{1}{n}$ and $\frac{1}{n^2}$ tend to 0. So the limit simplifies to:

   $\lim_{n \to \infty} \frac{25 - \frac{12}{n} + \frac{5}{n^2}}{1 - \frac{4}{n} + \frac{1}{n^2}} = \frac{25}{1} = 25$

Thus, the value of the limit is:

$\boxed{25}$

Would you like any further details or explanations on this solution?

Here are some related questions to expand on this concept:

1. What is the general technique for solving limits involving rational functions at infinity?
2. How does polynomial long division help in solving limits like this?
3. Can L'Hopital's Rule be used here, and why or why not?
4. What role does the degree of the polynomials play in determining the limit?
5. How do limits involving infinity relate to asymptotic behavior?

Tip: When dealing with limits of rational functions, simplifying by dividing both the numerator and denominator by the highest power of $n$ can make evaluating the limit much easier.