Limit with Floor Function II

Let $\sqrt{x} = \lfloor \sqrt{x}\rfloor +a$ , where $0\leq a<1$ , we have $\begin{array}{c}~ \large \lim_{x\to \infty}\dfrac{\lfloor \sqrt x \rfloor^2}{x} &= \lim_{x\to \infty}\dfrac{(\sqrt{x}-a)^2}{x}\\~\\&= \lim_{x\to \infty}\dfrac{x-2a\sqrt{x}+a^2}{x}\\~\\&=\lim_{x\to \infty}\dfrac{x}{x}-\dfrac{2a\sqrt{x}+a^2}{x}\\~\\&=\lim_{x\to \infty} 1-\dfrac{2a\sqrt{x}+a^2}{x}\\~\\&= 1-0\\~\\&=\boxed{1} \end{array}$

Let $\sqrt{x} = u$ and we know $\lfloor u \rfloor = u - r(u)$ , where $r(u) \in [0, 1)$ . Then

$\large \lim_{x\to \infty} \frac{ \lfloor \sqrt{x} \rfloor^2 }{x} = \large \lim_{u\to \infty} \left ( \frac{ \lfloor u \rfloor }{u} \right)^2 = \large \lim_{u\to \infty} \left( 1 - \frac{r(u)}{u} \right)^2 = 1,$

since $r(u)/u \rightarrow 0$ .

I did this very simply; hope I haven't overlooked anything. I simply observed that the square root of x, squared, equals x. Therefore the value will be 1 for any x.