Question: We proved by definition that the sequence $a\_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.

Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.

Then for all $n \geq N$ it holds, that

$$|a\_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{2n^3-8}{n^3-4} \big | = \big | \frac{n^2+11}{n^3-4} \big | = \frac{n^2+11}{n^3-4} \leq \frac{n^2+11}{n^3 - \frac{1}{2}n^3} \leq \frac{12n^2}{\frac{1}{2} n^3} = \frac{24}{n} < \epsilon $$

What I don't understand is how we get to $$- \frac{2n^3-8}{n^3-4} $$

Could someone explain? Thanks

Answer: 
Since $A\subset B$, we obtain: $$P(A\cap B)=P(A).$$
Id est, $$P(A^c\cap B)=\frac{1}{2}.$$
**Hint:**

$A\subset B$ so $A\cap B=A.$