Assistance in finding $\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}$ I am trying to prove $$\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n} = 0$$ given $n>0$.
But I'm having difficulties dealing with the floor function.  I tried splitting apart the limit like so:
\begin{align*}
\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n} &= 
\frac{\lim_{n\to\infty} (n-\lfloor \sqrt n \rfloor^2)}{\lim_{n\to\infty} n} \\
&= \frac{\lim_{n\to\infty} n}{\lim_{n\to\infty} n} - \frac{\lim_{n\to\infty} \lfloor \sqrt n \rfloor^2}{\lim_{n\to\infty} n}
\end{align*}
Trivially $\lim_{n \to \infty} \lfloor \sqrt n \rfloor^2 = \infty$, but it seems to grow somewhat slower than $\lim_{n \to \infty} n$ and so I am not sure if it is correct to conclude that $$\frac{\lim_{n\to\infty} \lfloor \sqrt n \rfloor^2}{\lim_{n\to\infty} n} = 1.$$

Using
$$x-1<\lfloor x \rfloor \le x$$
$$\lim_{n\to\infty} \frac{n-(\sqrt n)^2}{n}\le\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}\le\lim_{n\to\infty} \frac{n-(\sqrt n-1)^2}{n}$$
$$0\le\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}\le\lim_{n\to\infty} \frac{2\sqrt n-1}{n}=0$$
