How to prove this limit? Another answer is not helpful to me. I would've asked this in another post in the form of a comment, but I have too low a reputation. So therefore I'm forced to essentially ask a question again:
Prove that $\lim_{x\to\infty} \frac{x^{2}}{x-1}=\infty$ using the $\epsilon-M$ definition. In another post it was said that $\forall\epsilon>0\exists M>1 \text{ such that }M>x>1\rightarrow \frac{x^2}{x-1}>\epsilon$. I don't understand the condition on M. Say I choose $\epsilon$ to be $100$. Then if $x$ happens to be $3$ this condition is obviously broken no matter what $M$ I find. Shouldn't the condition on $M$ be as follows? $\exists M \text{ such that } x>M$ and then the usual pattern follows? In both possibilities I need help.
Thanks
 A: I'm going to make one more symbol change: for the same reason that $\delta$ (a small number) should be changed to $M$ (a large number), I'm going to change $\epsilon$ (also a small number) to $B$ (also a large number).
So to prove that $\lim_{x \to \infty} \frac{x^2}{x-1} = \infty$, you must prove:

For every $B > 0$ there exists $M > 0$ such that if $x > M$ then $\frac{x^2}{x-1} > B$.

Translating that logic into a proof strategy, here's what you actually have to do

Given $B > 0$ you must find an appropriate $M > 0$, and using that $M$ you must prove:
IF $x > M$ THEN $\frac{x^2}{x-1} > B$.

What happens at a particular value of $x$, such as $x=3$, is not at issue.
What is at issue is: given $B$, how do you find $M$?
The idea is that you start with the inequality that you are attempting to prove, in this case
$$(*) \qquad \frac{x^2}{x-1} > B
$$
and you attempt to "solve" this inequality for $x$, i.e. you attempt to find an $M$ such that the interval $x>M$ is in the solution set of $(*)$.
Once you think you've found the appropriate value of $M$, then the final steps of the logic are to prove that the interval $x > M$ is in the solution set of $(*)$. That is you must prove:

If $x > M$ then $\frac{x^2}{x-1} > B$.

It can be tricky to find the appropriate $M$. If you know how to solve quadratic inequalities then inequality $(*)$ is not too hard to solve. But that's a lot of work. Keeping in mind that you just have to find one $M$ that works, here's the one that I found, after poking around for a while, digging in the dirt, and clearing away the brush:

Let $M = \max\{B,1\}$

Actually, it's not all that hard to find that $M$, and perhaps you'll be able to tell how I found it. (I'll give my secret away at the end)
But anyway, now that I've found $M$, the final logic (one more time) is to prove:

If $x > M$ then $\frac{x^2}{x-1} > B$.

So, assume that $x > M = \max\{B,1\}$. It follows that $x > B$ and $x > 1$.
Since $x > 1$ it follows that $x-1 > 0$, and together with $x > x-1$ it follows that $\frac{x}{x-1} > 1$.
Therefore,
$$\frac{x^2}{x-1} = x \cdot \frac{x}{x-1} > B \cdot 1 = B
$$
and the proof is complete.

Perhaps you can back engineer my procedure for finding $M$ at this stage. Here's what I did:

I stared at the inequality $\frac{x^2}{x-1} > B$, and then broke up the left hand side into a product $x \cdot \frac{x}{x-1} > B$. Then I said to myself: "If I could make the first factor $>B$ and the second factor $>1$, I'd be done!" So all that was left was to find an $M$ so that $x > M$ implies that $x>B$ and $\frac{x}{x-1} > 1$.

A: What you want to show, in words, is that no matter how large you want $\frac{x^2}{x-1}$ to be, if you choose $x$ large enough, it will be. So in terms of a statement to prove, you want to show that given any $M>0$, there is some $N>0$ such that if $x>N$, then $\frac{x^2}{x-1}>M$.
A: You can write
$\tag 1 \displaystyle \frac{x^2}{x-1} = \frac{x^2- 1 + 1}{x-1} = \frac{x^2- 1}{x-1} + \frac{1}{x-1} =$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \displaystyle \frac{(x+1)(x-1)}{x-1} + \frac{1}{x-1} = x + 1 + \frac{1}{x-1}$
It is easy to conclude, by looking at (far) $\text{rhs}$ of $\text{(1)}$, that 
$\tag 2 \displaystyle \text{If } x \gt 1 \text{ then } \frac{x^2}{x-1} \gt x$
and that can used to 'prove the limit' for the OP.
A: You can't prove $\lim_{x\to \infty} f(x) = \infty$ with an $\epsilon-M$ proof.
$f(x)$ isn't approaching any finite value $L$ so we can't get $f(x)$ within some $\epsilon$ of $L$.  And getting $f(x)$ within $\epsilon$ of $\infty$ just doesn't make any sense.  (And not in the way "$f(x)$ is close to infinity" makes no sense; in the way "purple $f(x)$ to pancakes flap freckle $\infty$ the" makes no sense.  The words just don't mean anything in that order.)
There are four ways limits can behave and similar types of proofs.
1) $\lim_{x\to c} f(x) = L$.  Here $x$ "tends to" $c$-- a real number with a finite value$c$--- while $f(x)$ "tends to" $L$, also a real number.
The definition in this case is:

For every possible $\epsilon > 0$ we can find a $\delta > 0$ so that whenever $c-\delta < x < c+\delta$ we will have $L-\epsilon < f(x) < L + \epsilon$.

That's a $\delta-\epsilon$ proof.
2) $\lim_{x\to \infty} f(x) = L$.  In this case $x$ doesn't tend to a value, $x$ grows unbounded and while it does so $f(x)$ "nears" the value $L$.
The definition for this is:

For every possible  $\epsilon$ we can find an $N$ so that if $x > N$ then we will always have $L-\epsilon < x < L+\epsilon$.

In this case we don't try to find a $\delta$ because we aren't looking at $x$ being within $\delta$ of a value.  We have $x$ being unbounded large so we looking at $x$ being large enough that it is larger than something.
Hence an "$\epsilon-N$ proof.
(Note:  We can define $\lim_{x\to -\inf}f(x)$ the same way.  But not we are looking for an $N$ where whenever $x < N$ then $L-\epsilon < f(x) < L + \epsilon$
3) $\lim_{x\to c} f(x)= \infty$.
In this case we have $x$ "approaching" a value $c$ but now it is $f(x)$ that is growing unbounded.
(Geometrically, $x = c$ would be a vertical assymptote.)
So the definitions is:

For any possible $M$ we can find a $\delta > 0$ so that whenever we have $c-\delta < x < c+\delta$ we will have $f(x) > M$.

The idea is no matter how big $M$ is we can find an $f(x)$ that is even larger by picking $x$ very close, withing $\delta$, of c.
A $\delta-M$ proof is nesc.
4)  $\lim_{x\to \infty} f(x) = \infty$.
In this case both $x$ grows unbounded and "as" it does so, $f(x)$ grows unbounded.  There is no value either of them are getting "close" to and so there is no range we want to put them in.
The definition in this case is:

For an $M$ there will be an $N$ so that whenever $x > N$ then $f(x) > M$.

We could call this and "$M-N$ proof but I've never heard anyone ever do that.
....
Note all four of the are similar.  In all of them we want to force a condition on $f(x)$,  either that $f(x)$ is within a small $\epsilon$ distance of a limit value, $L$ or that that $f(x)$ is larger than some large value $M$.  And we force this condition but choosing an $x$ with conditions; either that $x$ is within a small distance, $\delta$ of a target point $c$, or that $x$ has grown larger than some large value $N$.
.....
So in this case we have case 4)
We want to prove that $\lim_{x\to \infty} \frac {x^2}{x-1} = \infty$.
So we want to show: for any $M$, no matter how large, we can always have $\frac {x^2}{x-1} > M$ by taking large values of $x$...  The question is... How large?  What $N$ must we have so that picking $x> N$ will always result in $\frac {x^2}{x-1} > M$.
Well, there are several ways to do this.  The easiest that comes to my mind is that:
As long as we picking large values of $x$ we can assume we are picking $x > 1$ and so $0 < x -1 < x$ so $x= \frac {x^2}{x} < \frac {x^2}{x-1}$.
So we can pick $N = \max(M, 1)$.
If $x > N$ then $x > M$ and $\frac {x^2}{x-1} > x > M$.  And that's it.  We are done.
