If $\lim_{n \to \infty}x_n=x$ , then $\lim_{n \to \infty} \left(\lfloor x_n^2\rfloor+\lfloor x_n\rfloor\right)$ =?

Question

If $\lim_{n \to \infty}x_n=x,$ where $x$ is any real number,
then $\lim_{n \to \infty}\left(\lfloor x_n^2\rfloor+\lfloor x_n\rfloor\right) =?,$ where $\lfloor y\rfloor$ denotes the greatest integer less than or equal to $y.$

My approach: if $x\notin \Bbb Z$ ,then $\lfloor\cdot\rfloor$ is continuous at x.
So $\lim_{n \to \infty}\left(\lfloor x_n^2\rfloor+\lfloor x_n\rfloor\right)=x^2+x$
Now if $x\in\Bbb Z$ then
$Case 1$: if ${x_n}$ takes value from the left side then, $\lfloor x_n\rfloor$ converges to $x-1$
So in this case $\lim_{n \to \infty}\left(\lfloor x_n^2\rfloor+\lfloor x_n\rfloor\right)=x^2+x-1$
$Case 2$: if $x_n$ takes value from the right side then, $[x_n]$ converges to $x$
So in this case $\lim_{n \to \infty}\left(\lfloor x_n^2\rfloor+[x_n\rfloor\right)=x^2+x$
Is my approach correct?

Answer 1

As the statement is currently written, I think your statement is untrue. Take $x_n= m +\frac{(-1)^n}{n}$ for some $m\in \mathbb{Z}$. Then $[x_n]=m$ for $n$ even and $[x_n]=m-1$ for $n$ odd. Likewise, for $n> 4\vert m\vert$ we have $[x_n^2]=z^2$ when $n$ is even or $[x_n]=z^2-1$ when $n$ is odd.

If you assume monotonicity and $x\in \mathbb{Z}$, then there exists some $N_0$ such that $x_n^2\in [x^2-1,x^2)$ or $x_n^2=x^2$ or $x_n^2\in [x^2,x^2+1)$. In all these cases, for $n>N_0$ you have that $[x_n^2]+[x_n]$ is equal to the left hand sides of the equalities you wrote.