Binomial theorem used in real analysis to prove limits So we have $\lim\limits_{n \to \infty} n^{1/n}=1$
$$|n^{1/n}-1|<\epsilon \Rightarrow n<(\epsilon+1)^n $$
Then using binomial theorem:
$$n<1+n\epsilon+\frac{n(n-1)}{2}\epsilon^2+...+\epsilon^n \ge \frac{n(n-1)}{2}\epsilon^2$$
we can conclude $$n<\frac{n(n-1)}{2}\epsilon^2$$
Then just some algebra to get the N and the usual stuff.
My question is how do we know that $n<\frac{n(n-1)}{2}\epsilon^2$? Can't it be that $n>\frac{n(n-1)}{2}\epsilon^2$?
 A: You can't conclude this, but this is not a problem. Remember the implication at the heart of the limit definition:
$$n > N \implies |n^{1/n} - 1| < \varepsilon. \tag{$\star$}$$
Note the direction of the arrow $\implies$. This means that all of our logic has to point towards the conclusion $|n^{1/n} - 1| < \varepsilon$.
Normally, in a proof, we start at the premises, and we want to write a sequence of statements, where every statement is implied by the statement(s) before. We begin at the premises, and end at the conclusion.
Here, it's a bit different. Often when working out limit definition arguments, we are starting at the conclusion $|n^{1/n} - 1| < \varepsilon$ and working backwards towards the premise (because we don't yet know what $N$ will work!). So, in this case, all statements should imply the previous statements. This is a bit of an adjustment, but it's necessary so we can find a suitable $N$. Once we find such an $N$, then reversing the steps will provide a logical proof of $(\star)$. Now, all statements will imply the next statement, thus making a logical proof, starting from premise, and ending at conclusion (now that we actually know the premise, it's not so bad!).
In this case, as you point out, we do not have
$$n<1+n\varepsilon+\frac{n(n-1)}{2}\varepsilon^2+...+\varepsilon^n \implies n < \frac{n(n-1)}{2}\varepsilon^2.$$
However, given we are working backwards, this implication is not relevant! The important implication is
$$n < \frac{n(n-1)}{2}\varepsilon^2 \implies n<1+n\varepsilon+\frac{n(n-1)}{2}\varepsilon^2+...+\varepsilon^n,$$
which is absolutely true, given that the remaining summands are all non-negative. This is important, as the choice of $N$ that you eventually choose will satisfy
$$n > N \implies n < \frac{n(n-1)}{2}\varepsilon^2,$$
which then allows you to logically conclude
$$n<1+n\varepsilon+\frac{n(n-1)}{2}\varepsilon^2+...+\varepsilon^n,$$
and finally $|n^{1/n} - 1| < \varepsilon$.
A: I think it can be. For instance $$1>\frac{1(0)}{2}\epsilon^2$$ for any positive epsilon. What seems to be missing is some mention in your text of "for all big enough" $n$.
Also notice that if $n=2$, say, then $n<(1+\epsilon)^n$ will fail for certain epsilon.
A: When you write
$$|n^{1/n}-1| < \epsilon$$ what you are really writing is the disk
$$-\epsilon < n^{1/n}-1 < +\epsilon$$
The fact that $n\to +\infty$ allows you to remove the absolute value since it's always positive, hence you get
$$n < (\epsilon +1)^n$$ as you stated.
Now by the definition, in this case, you need to find $M_n$ such that $\forall x > M_n$ it holds $|f(x) - \ell| < \epsilon$
What you perhaps are missing is to remember that here $n\to +\infty$, big enough.
Later with a bit of algebra, you can recover $n$ from the last passage of yours:
$$n < \frac{n^2-n}{2}\epsilon^2 \longrightarrow n > \frac{2}{\epsilon^2} + 1$$
Which as $\epsilon \to 0$ is big enough as wanted, and we got our $M_n$.
