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

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?

• Is the curly brackets meant to be the fractional part? Or just normal brackets? Commented Sep 3 at 11:48
• normal brackets Commented Sep 3 at 11:50
• Are you assuming that $x_n$ is monotonic? Commented Sep 3 at 11:52
• yes but if $x_n$ is not monotonic ,is this affect the limit points? Commented Sep 3 at 11:58

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.