To prove that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$, we need to prove both directions of the statement:

$(\Rightarrow)$ Assume $a_0 = \sup A$. Let $\epsilon>0$ be given. Since $a_0$ is the least upper bound of $A$, there exists $a\in A$ such that $a_0-\epsilon<a\leq a_0$. This is because if there were no such $a$, then $a_0-\epsilon$ would be a smaller upper bound for $A$ than $a_0$, contradicting the assumption that $a_0=\sup A$.

$(\Leftarrow)$ Assume that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. We need to show that $a_0=\sup A$. Suppose, for the sake of contradiction, that $a_0$ is not the least upper bound of $A$. Then there exists $b>a_0$ such that $b$ is an upper bound for $A$. Let $\epsilon = b-a_0>0$. Since $a_0$ is not the least upper bound of $A$, there exists $a\in A$ such that $a>b$. But then $a_0< a< b = a_0 + \epsilon$, which contradicts the assumption that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. Therefore, $a_0$ is the least upper bound of $A$, i.e., $a_0=\sup A$.

Hence, we have proven that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$.

We can prove this limit using the squeeze theorem. First, note that for any positive $₫ \mathcal{P}$ integer \$ n \$, we have:
$$
20 n^2+20 n+2020 \geq 20 n^2
$$
This is because \$ 20 n+2020 \$ is always positive for positive \$ n \$, so adding it to \$ 20 n^{\wedge} 2 \$only increases the value. Using this inequality, we can write:
$$
0 \leq \frac{1}{20 n^2+20 n+2020} \leq \frac{1}{20 n^2}
$$
Now, taking the limit as n approaches infinity of both sides of this inequality, we get:
$$
\lim {n \rightarrow \infty} 0 \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020} \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2} $$ The left-hand side simplifies to 0 , and the right-hand side simplifies to 0 as well, since the denominator grows faster than the numerator as $\$ \mathrm{n} \$$ approaches infinity. Therefore, by the squeeze theorem, we have: $$ \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020}=0
$$
as desired.

First, we will prove that if $\lim_{n \rightarrow \infty} x_n = 0$, then $\limsup_{n \rightarrow \infty} |x_n| = 0$.

Suppose $\lim_{n \rightarrow \infty} x_n = 0$. Let $\epsilon > 0$ be arbitrary. Since $\lim_{n \rightarrow \infty} x_n = 0$, there exists $N$ such that for all $n \geq N$, we have $|x_n| < \epsilon$.

Thus, for any $k \in \mathbb{N}$, there are only finitely many $n$ such that $|x_n| \geq \frac{1}{k}$. Therefore, $\limsup_{n \rightarrow \infty} |x_n| = 0$.

Next, we will prove that if $\limsup_{n \rightarrow \infty} |x_n| = 0$, then $\lim_{n \rightarrow \infty} x_n = 0$.

Suppose $\limsup_{n \rightarrow \infty} |x_n| = 0$. Then, for any $\epsilon > 0$, there exists $N$ such that for all $n \geq N$, we have $|x_n| < \epsilon$.

Since ${x_n}$ is bounded, there exists a positive number $M$ such that $|x_n| \leq M$ for all $n$. Therefore, for any $\epsilon > 0$, we have

$\left|x_n\right| \leq M<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

for all $n \geq N$. This implies that $\lim_{n \rightarrow \infty} x_n = 0$.

Thus, we have shown both directions of the equivalence and conclude that

$\lim {n \rightarrow \infty} x_n=0 \quad$ if and only if $\quad \limsup {n \rightarrow \infty}\left|x_n\right|=0$.