The expression given is:

\[
\lim_{{y \to 2}} \frac{y^3 - y^2 - y - 2}{2y^2 - 5y + 6}
\]

To find this limit as \( y \) approaches 2, we'll first try to evaluate the expression directly by substituting \( y = 2 \). If direct substitution results in an indeterminate form (like \( \frac{0}{0} \)), we'll need to simplify the expression by factoring or using other techniques.

### Step 1: Direct Substitution
Substituting \( y = 2 \) into the expression:

For the numerator:
\[
y^3 - y^2 - y - 2 = 2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0
\]

For the denominator:
\[
2y^2 - 5y + 6 = 2(2)^2 - 5(2) + 6 = 8 - 10 + 6 = 4
\]

Thus, the expression becomes:
\[
\frac{0}{4} = 0
\]

### Final Answer:
Since the limit does not result in an indeterminate form, the limit is simply:

\[
\lim_{{y \to 2}} \frac{y^3 - y^2 - y - 2}{2y^2 - 5y + 6} = 0
\]

Would you like further details, or do you have any questions?

Here are five related questions:
1. How do you factor polynomials to simplify limits?
2. What is an indeterminate form, and how do you resolve it?
3. How do you apply L'Hôpital's rule in limits involving indeterminate forms?
4. What are common techniques for finding limits in calculus?
5. How do limits relate to the continuity of a function?

**Tip:** Always check if direct substitution is possible in limit problems before trying more complex methods.

Learn how to compute the limit of (y^3 - y^2 - y - 2) / (2y^2 - 5y + 6) as y approaches 2 with a step-by-step solution. This problem involves polynomial functions and direct substitution technique in calculus. Suitable for advanced high school students (Grades 11-12).

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