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

Question

Let $X_n$ be a sequence of non negative real numbers. Then which of the following is true?

1. $\liminf_{n\to\infty} X_n = 0 \implies \lim_{n\to\infty} X_n^2 = 0$
2. $\limsup_{n\to\infty} X_n = 0 \implies \lim_{n\to\infty} X_n^2 = 0$
3. $\liminf_{n\to\infty} X_n = 0 \implies X_n$ is bounded
4. $\liminf_{n\to\infty} X_n^2 > 4 \implies \limsup_{n\to\infty} X_n > 4$

Please Help me to Prove or Disprove the given options.

My Attempt:

Option 1 can be discarded by taking the sequence (0,1,0,1,0,1,0,1,...)

Option 3 can be discarded by taking the sequence (0,1,0,2,0,3,0,4,0,5,0,6,...)

Option 4 can be discarded by taking the sequence (3,3,3,3,3,...)

Please help me to Prove option 2.

Answer 1

If $\limsup_{n\to\infty}x_n=0$, then there exists $N$ such that $\sup_{n\geqslant N}x_n<1$, and hence $x_n^2<x_n$ for $n\geqslant N$. It follows then $$\limsup_{n\to\infty} x_n^2\leqslant \limsup_{n\to\infty} x_n=0,$$ and from $x_n\geqslant 0$ that $\liminf_{n\to\infty} x_n^2\geqslant 0$. Since trivially $\liminf_{n\to\infty} x_n^2\leqslant\limsup_{n\to\infty} x_n^2$, we find that $$ \liminf_{n\to\infty}x_n^2 = \limsup_{n\to\infty} x_n^2 = 0, $$ and hence $\lim_{n\to\infty} x_n^2=0$.