Here is the analysis and solution to the given mathematical problems. Let's break the exercises into parts for clarity.

Exercise 01

Calculate the following limits, justifying your answers:

a)

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

• The term of highest degree is $-2n^3$, as it dominates all others for large $n$.
• Therefore: $\lim_{n \to +\infty} \left( -2n^3 - 5n^2 + 4n - 2 \right) = -\infty$

b)

$\lim_{n \to +\infty} \frac{n^2 - 4n}{5n^3 + 2n^2 + 1}$

• The highest degree in the numerator is $n^2$, and in the denominator is $5n^3$.
• Dividing through by $n^3$: $\lim_{n \to +\infty} \frac{\frac{n^2}{n^3} - \frac{4n}{n^3}}{\frac{5n^3}{n^3} + \frac{2n^2}{n^3} + \frac{1}{n^3}} = \lim_{n \to +\infty} \frac{\frac{1}{n} - \frac{4}{n^2}}{5 + \frac{2}{n} + \frac{1}{n^3}}$
• As $n \to +\infty$, all terms involving $\frac{1}{n}, \frac{1}{n^2},$ or $\frac{1}{n^3}$ approach 0.
• Result: $\lim_{n \to +\infty} \frac{n^2 - 4n}{5n^3 + 2n^2 + 1} = 0$

c)

$\lim_{n \to +\infty} \sqrt[3]{n^2 - n + 1}$

• The dominant term inside the cubic root is $n^2$.
• Therefore: [ \lim_{n \to +\infty} \sqrt[3]{n^2 - n + 1} = \sqrt[3]{n^2} = n^{2/3}. ]

d)

$\lim_{n \to +\infty} \frac{3 + 3\sin(n^2 + n)}{n + 5}$

• The denominator grows unbounded as $n \to +\infty$, while the numerator oscillates between $3 - 3 = 0$ and $3 + 3 = 6$.
• The overall limit approaches 0 because the denominator dominates: $\lim_{n \to +\infty} \frac{3 + 3\sin(n^2 + n)}{n + 5} = 0.$

Exercise 02

a)

Show that for all $n \in \mathbb{N}$: $-3 - \frac{3}{n+5} \leq -3 + \frac{3\sin(n^2 + n)}{n+5} \leq -3 + \frac{3}{n+5}.$

Proof:

• The sine function satisfies: $-1 \leq \sin(n^2 + n) \leq 1.$
• Multiplying by $\frac{3}{n+5}$ (a positive value for $n \geq 0$): $-\frac{3}{n+5} \leq \frac{3\sin(n^2 + n)}{n+5} \leq \frac{3}{n+5}.$
• Adding $-3$ to all sides: $-3 - \frac{3}{n+5} \leq -3 + \frac{3\sin(n^2 + n)}{n+5} \leq -3 + \frac{3}{n+5}.$

b)

Deduce by the squeeze theorem: $\lim_{n \to +\infty} \left( -3 + \frac{3\sin(n^2 + n)}{n+5} \right) = -3.$

Reasoning:

• From part (a), as $n \to +\infty$, the bounds $-\frac{3}{n+5}$ and $\frac{3}{n+5}$ both approach 0.
• By the squeeze theorem: $\lim_{n \to +\infty} \left( -3 + \frac{3\sin(n^2 + n)}{n+5} \right) = -3 + 0 = -3.$

Do you want me to provide further clarifications?

Here are 5 related questions to deepen your understanding:

1. How does the squeeze theorem work, and why is it useful in limits?
2. Why does the dominant term of a polynomial determine the behavior of a limit as $n \to \infty$?
3. Can the sine function affect the limit result significantly when it is divided by a term growing unbounded?
4. What is the importance of dividing numerator and denominator by the highest power of $n$ in rational expressions?
5. How would the results change if these limits were computed as $n \to -\infty$?

Tip: For polynomial limits, always simplify by focusing on the highest degree term—it dominates the behavior at infinity.