The limit $\lim\limits_{n\to\infty} (\sqrt{n^2-n}-n)$. Algebraic and intuitive thoughts. I am working on the following problem.

Find the limit of $$\lim_{n \to \infty} (\sqrt{n^2-n}-n)$$

Intuitively, I want to say it's $0$ because as $n \to \infty$, $\sqrt{n^2-n}$ behaves like $n$ and subtracting $n$ makes it $0$.
However, algebraically, to my surprise
$$\begin{align}
\lim_{n \to \infty} (\sqrt{n^2-n}-n) & = \lim_{n \to \infty} \frac{n}{\sqrt{n^2-n}+n}\\
& = \lim_{n \to \infty} \frac{1}{\sqrt{1-\frac{1}{n}}+1}\\
& = \frac{1}{2} \\
\end{align}$$
Is there any intuitive way to explain why 0 was not the answer ?
 A: Using $(1+x)^\alpha \approx_0 1+\alpha x$ we have:
$$\sqrt{n^2-n}-n=n\left(\sqrt{1-\frac{1}{n}}-1\right)\approx_\infty n\left(1-\frac{1}{2n}-1\right)=-\frac{1}{2}$$
A: You are wrong, because the expression is clearly negative for $n \gt 0$
One way of seeing the result is to note that $n^2-n=\left(n-\frac 12\right)^2-\frac 14$. As $n$ increases, the $\frac 14$ becomes insignificant.
You should get $-\frac 12$
A: If you know that $\sqrt{1+x}\sim1+\frac x2$ for $x\sim0$, we get that
$$
\begin{align}
\lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right)
&=\lim_{n\to\infty}n\left(\sqrt{1-\frac1n}-1\right)\\
&=\lim_{n\to\infty}n\left(\left(1-\frac1{2n}\right)-1\right)\\
&=\lim_{n\to\infty}n\left(-\frac1{2n}\right)\\
&=-\frac12
\end{align}
$$
or to complete your answer and correct the sign:
$$
\begin{align}
\lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right)
&=\lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right)\frac{\sqrt{n^2-n}+n}{\sqrt{n^2-n}+n}\\
&=\lim_{n\to\infty}\frac{(n^2-n)-n^2}{\sqrt{n^2-n}+n}\\
&=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}\\
&=\lim_{n\to\infty}\frac{-1}{\sqrt{1-1/n}+1}\\
&=-\frac12
\end{align}
$$
A: You have a sign error: you should have $-\frac12$.
Complete the square: $n^2-n=\left(n-\frac12\right)^2-\frac14$, so for large $n$ its square root is very close to $n-\frac12$, and subtracting $n$ brings you very close to $-\frac12$.
