I have to evaluate lim n → ∞ ( − 1 ) n − 1...

The problem is that as $n\to \mathrm{\infty }$, $\sqrt{n^2+0.5n+1}$ diverges, and $\sqrt{n^2+0.5n+1}-n\ne 0$
. We have
$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\sqrt{n^2+0.5n+1}-n& =\underset{n\to \mathrm{\infty }}{lim}\sqrt{(n+0.25{)}^2+1-{0.25}^2}-n\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{(\sqrt{n^2+0.5n+1}-n)(\sqrt{n^2+0.5n+1}+n)}{\sqrt{n^2+0.5n+1}+n}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{0.5n+1}{\sqrt{n^2+0.5n+1}+n}=\frac{1}{4}\end{array}$

Before $1+\frac{0.5}{n}+\frac{1}{n^2}$ can "become" 1, it is greater than 1. And so when it multiplies by $n^2$, you get something greater than $n^2$. And it turns out that it is still greater enough than $n^2$ to make a difference.
Use $n^2+0.5n+1=(n+0.25{)}^2+\frac{15}{16}$. Then you have $(n+0.25)\sqrt{1+\frac{15}{16(n+0.25{)}^2}}$. At this point your intuition works out to be correct, and $1+\frac{15}{16(n+0.25{)}^2}$ goes to 1 quickly enough. But more should be done to prove this.