Proving $\lim_{x\to -\infty} \frac{x^2}{x+1}=-\infty$ I have not been able to find any examples for this, so I'm unsure about my presentation (or answer for that matter). If anyone could also provide a link to a list of examples that I could work through, that would be very much appreciated. 
Question: 
Use the precise definition to prove:-
$$\lim_{x\to -\infty} \frac{x^2}{x+1}=-\infty$$
Attempted answer:
For every number $N < 0$ there is some number $M < 0$ such that:
-->   $- f(x) < N$ whenever $-x < M$
-->   $-|\frac{x^2}{x+1}|< N$ 
-->   $-|{x}| < -|\frac{x^2}{x+1}| < N$
-->   $-|x| < N$
-->   Let M = N, then $-x < M$ implies that $-f(x) < N$
This is all over the place. Help would be appreciated. Thanks. 
 A: We may suppose $N <-2$. You want $\frac {x^{2}} {x+1} <N$. This is same as $x^{2} >N(x+1)$ (when $x+1 <0$). Write this as $(x-\frac N 2)^{2}>N+\frac {N^{2}} 2$. This in equality holds if $x <\frac N 2 - \sqrt { N+\frac {N^{2}} 2}$. So take $M=\frac N 2 - \sqrt { N+\frac {N^{2}} 2}$. This gives $f(x) <N$ whenever $x <M$. 
A: This certainly is all over the place, but that's nothing surprising for a novice in the field. Writing mathematical proofs is not easy, and it takes skill, patience and practice.
What you wrote is the process one should take when trying to prove a statement, but 1) it is wrong, because $-|x|<\left|\frac{x^2}{x+1}\right|$ is false for $x=10$, for example, and 2) you are missing one final part of that, and that is, once you see where you are, you need to actually write the proof.
That is, once you discover what $M$ should be, then take that $N$ and write the proof in the following form:

Let $N>0$. Then, take $M=...$ and let $x < M$. Then, $\dots$, and therefore, $\frac{x^2}{x+1} < N$.

where of course, you fill in the places where "$\dots$" is written
