Prove by definition that: $\displaystyle\lim_{n\to\infty}\frac {1-n^2}{2n^2+n+5}=-\frac 1 2$
Scratch work to find $N\in \mathbb R$: $|\frac {1-n^2}{2n^2+n+5}+\frac 1 2|=\frac {n+7}{4n^2+2n+10}\color{red}{<}\frac {n+7}{n^2}=\frac 1 n+\frac 7 {n^2}<\epsilon$ so take $N=max\{\frac 1 {\sqrt\epsilon},\frac 7 \epsilon\}$.
But this $N$ leads me to this expression: $\epsilon +7\sqrt\epsilon$ which I don't think is correct. I'm not sure the red $<$ is correct but I don't see any other way to make the denominator smaller...
Note the polynomial in the denominator has no solutions, so it can't be simplified.
The red inequality might or might not always true, I don't know, didn't check (edit: it's obviously true because $n$ is always positive). But it will be true for big enough $n$ and that's good enough. Proceed with $\dfrac 1 n+\dfrac 7 {n^2}\leq \dfrac 1 n+\dfrac 7 n=\dfrac 8 n$.