Proof using formal definition: Infinite limit I was wondering how get the proof of this limit:
$$\lim\limits_{x\to -\infty}\dfrac{{x^2} - x + 1}{x + 4} = -\infty$$
The problem is that I don't know what to do for find the appropriated values to make valid the implication of the formal definition (epsilon-delta).
I would appreciate if somebody can help me. 

You want to find $N(M)<0$ (a function in terms of $M<0$) such that
$$x<N(M)\implies \frac{x^2-x+1}{x+4}<M$$
Let $N(M)\le -4$. Then $x<-4$ and $$\frac{x^2-x+1}{x+4}<M\iff x^2-x+1>M(x+4)$$
$$\iff x^2-x(M+1)+(1-4M)>0$$
If $M\le -9-2\sqrt{21}$, then $\Delta=M^2+18M-3\ge 0$ and let $$N(M)=\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}$$
If $M\in(-9-2\sqrt{21},0)$, then let $$N(M)=\min\left\{-4,\frac{k+1-\sqrt{k^2+18k-3}}{2}\right\}$$
for any $k\le-9-2\sqrt{21}$, e.g. you can let $k=-19$:
$$N(M)=\min\left\{-4,-11\right\}=-11$$
Answer: you can let $$N(M)=\begin{cases}\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}, && M\le -9-2\sqrt{21}\\-11, && M\in(-9-2\sqrt{21},0)\end{cases}$$
