To prove the given limit, we will focus on the expression:

$\lim_{n \to +\infty} \frac{3n^2 + 2n - 5}{2n^2 - 5} = \frac{3}{2}$

Step-by-Step Proof:

1. Focus on the highest powers of $n$: When $n$ approaches infinity, terms with the highest power of $n$ dominate. The highest power of $n$ in both the numerator and denominator is $n^2$.

2. Factor out $n^2$ from both the numerator and the denominator:

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

3. Simplify the expression by canceling out the $n^2$ terms:

   $= \frac{3 + \frac{2}{n} - \frac{5}{n^2}}{2 - \frac{5}{n^2}}$

4. Take the limit as $n \to \infty$: As $n$ becomes infinitely large, both $\frac{2}{n}$ and $\frac{5}{n^2}$ tend to 0. Thus, we are left with:

   $= \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2}$

Hence, the limit is indeed:

$\boxed{\frac{3}{2}}$

Conclusion:

The limit is correctly evaluated as $\frac{3}{2}$, and the steps confirm the given statement.

Would you like more details on any specific step?

Here are five related questions to consider:

1. What happens if the degree of the numerator is lower than that of the denominator?
2. How would you handle limits if the degree of the numerator were higher than the denominator?
3. How do we handle limits when there's an indeterminate form like $\frac{\infty}{\infty}$?
4. What is the general rule for limits of rational functions at infinity?
5. How can L'Hôpital's rule be applied to similar problems?

Tip: When calculating limits at infinity for rational functions, always compare the degrees of the numerator and denominator to determine the behavior.