Prove that for $ f: \mathbb R \to \mathbb R , f(x)=\frac {x-1} {x^2 + x + 2}, \lim _{x \to 0} f(x) =- \frac12 $ by epsilon delta

Question

As you can see from the image, I am not too sure where to go from after this. Could anyone give me a (not too ambiguous) hint?

Answer 1

Hint: Note that $x^2+x+2=\left(x+\frac12\right)^2+\frac74\geqslant\frac74$. On the other hand, you proved that $|x|<1\implies|x+3|<4$. Therefore$$|x|<1\implies\left|\frac{x(x+3)}{x^2+x+2}\right|<\frac{16}7|x|.$$