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. 
 A: 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 $$N(M)=\begin{cases}\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}, && M\le -9-2\sqrt{21}\\-11, && M\in(-9-2\sqrt{21},0)\end{cases}$$
A: Expand what you know (as one can tell from your comment below Kay K.'s deleted answer) to a "snapshot" argument is not difficult, in fact.
If $x < -4$, then
$$
\frac{x^{2}-x+1}{x+4} = x-5 + \frac{21}{x+4} < -9 + \frac{21}{x+4};
$$
given any $M < -9$, we have
$-9 + 21/(x+4) < M$ if in addition
$$
x < \frac{21}{M+9} - 4.
$$
A: Note that
$$\frac{x^2-x+1}{x+4}=\frac{(x+4)(x-5)+21}{x+4}.$$
You need to show that given $M<0$, there exists $N<0$ such that $\frac{x^2-x+1}{x+4}<M$ for all $x<N$. Let $N=\min\{M,-4\}.$ Then if $x<N$,
\begin{align*}
\frac{x^2-x+1}{x+4}&=\frac{(x+4)(x-5)+21}{x+4}\\
&=(x-5) + \frac{21}{x+4}\\
&<(N-5) + \frac{21}{x+4}\\
&<M+\frac{21}{x+4}\\
&<M.
\end{align*}
We can drop the $\frac{21}{x+4}$ because $x<N \leq -4$, so $x+4$ is negative, so $\frac{21}{x+4}<0$.
