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 $$\frac{\lim_{n\to\infty} \lfloor \sqrt n \rfloor^2}{\lim_{n\to\infty} n} = 1.$$

For very large values of $n$, the floor of $\sqrt{n}$ is almost indistinguishable (in the relative sense) from $\sqrt{n}$ itself; the absolute difference is at most $1$, so the relative difference is tending to zero.
To make this precise, note that we have the inequalities
$$\sqrt{n} - 1 \le \lfloor \sqrt{n} \rfloor \le \sqrt{n}$$
Thus upon squaring,
$$n - 2\sqrt{n} + 1 \le \lfloor \sqrt{n}\rfloor^2 \le n$$
After some algebraic work, you should be able to reduce the problem to computing $\lim_{n \to \infty} \sqrt{n} / n$.
