We are given the expression $\lim_{x \to 3} (x^2 - x) = 6$, and we want to evaluate whether the limit holds.

Step 1: Verify if $x^2 - x = 6$ when $x = 3$

Substitute $x = 3$ into the expression $x^2 - x$:

$x^2 - x = 3^2 - 3 = 9 - 3 = 6$

Indeed, $x^2 - x = 6$ when $x = 3$.

Step 2: Take the limit

Since $x^2 - x$ is a continuous function, we can directly substitute $x = 3$ into the expression:

$\lim_{x \to 3} (x^2 - x) = 3^2 - 3 = 9 - 3 = 6$

Thus, the limit is:

$\boxed{6}$

Let me know if you'd like further details or clarifications.

Here are 5 related questions to practice:

1. What is $\lim_{x \to 2} (x^2 - x)$?
2. How do you evaluate $\lim_{x \to 4} (x^2 - 2x + 1)$?
3. What happens if you evaluate $\lim_{x \to 0} (x^2 - x)$?
4. Find $\lim_{x \to 5} (x^2 - 2x)$.
5. How do you handle limits with discontinuous functions?

Tip: When evaluating limits for polynomial functions, substituting the value directly often works, as polynomials are continuous everywhere.