Let's solve the exercises shown in the new image step by step.

Exercise 2:

1. $\lim_{x \to +\infty} \frac{3x^3 - 4x^2 + 6}{3x^2 - 2x + x^3}$

Step 1: Simplify the expression.

• Both the numerator and denominator contain terms of $x^3$. The highest degree term in both is $x^3$.
• Divide both the numerator and denominator by $x^3$ (the highest degree term):

$\frac{3x^3 - 4x^2 + 6}{3x^2 - 2x + x^3} = \frac{3 - \frac{4}{x} + \frac{6}{x^3}}{1 + \frac{-2}{x^2} + \frac{1}{x^3}}$

Step 2: As $x \to +\infty$, the terms with powers of $x$ in the denominator (like $\frac{4}{x}$, $\frac{6}{x^3}$, etc.) vanish. So, we are left with:

$\frac{3}{1} = 3$

Thus, the limit is: $\lim_{x \to +\infty} \frac{3x^3 - 4x^2 + 6}{3x^2 - 2x + x^3} = 3$

Exercise 3:

2. $\lim_{x \to 2} \frac{x^3 + 2x^2 - 8x}{x^3 - 2x^2 + 2x - 4}$

Step 1: Factor both the numerator and denominator.