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