$\liminf x_n = 0 \implies \lim x_n^2 = 0$

Question

Let $(x_n)_{\geq1}$ be a sequence of non negative real numbers. Then which of the following statement is true?

1. $\liminf x_n = 0 \implies \lim x_n^2 = 0$

2. $\limsup x_n = 0 \implies \lim x_n^2 = 0$

3. $\liminf x_n = 0 \implies $$(x_n)_{\geq1}$ is bounded.

4. $\liminf x_n^2 > 4 \implies \limsup x_n > 4$

My Attempt:

We know that

1. $\lim_{x \to \infty} x_n = l \implies \liminf x_n = l = \limsup x_n$
2. If terms of the sequence $(x_n) > 0 \implies \lim_{x \to \infty} x_n = l \geq 0$
3. $\liminf x_n$ = smallest limit point. And $\limsup x_n $ = largest limit point. Counter example for option 3 is $(0,1,0,2,0,3,0,4,...)$ and Counter example for option 4 is $(4,4,4,...)$ and I have no counter example for 1. Also how to prove 2 by easiest approach.

Answer 1

For 1. it's a matter of definitions. The series $1,0,2,0,3,0,4\ldots$; i.e., $x_{2n}=0$ and $x_{2n+1} =n$ has a $\lim\inf$ of $0$ for each nonnegative integer $n$ but no limit. So 1. cannot be true.

For 2., simply note that $\lim \sup x_n = 0$ implies for all $\epsilon$ satisfying $1>\epsilon >0$ an integer $n_{\epsilon}$ s.t. $x_n < \epsilon$ for all $n \ge n_{\epsilon}$. Given that each such $x_n$ is nonnegative , what can you say about all $x^2_n$ for all $n \ge n_{\epsilon}$. Can't you say $x^2_{n} \le \epsilon^2 \le \epsilon$ for all such $n$ as well.