Proofs in Real Analysis are too 'convenient' I'm doing a first course in real analysis and I have studied nearly 10-15 theorems and proofs by now. One thing I've noticed in all of them is that they all seem too 'convenient' and full of assumptions. This, I find very peculiar to real analysis. To understand my point, consider this one for instance:

Theorem: Let $\{x_n\}$ be a sequence of $\mathbb R^+$ such that $\lim_{n\to\infty} |\frac{x_{n+1}}{x_n}| = l$. If $0\le l \lt 1$, $\lim_{n\to\infty } x_n = 0$.
Proof: Consider $\epsilon \gt 0$ such that $l+\epsilon \lt 1$. There exists $N \in \mathbb N$ such that $||\frac{x_{n+1}}{x_n}| - l |\lt \epsilon$ for all $n \ge N$.
$l-\epsilon \lt |\frac{x_{n+1}}{x_n}| \lt l+\epsilon$ (for all $n \ge N$.)
Let $m = l + \epsilon$. Given that $0 \le l \lt 1$, we could say that $0 \lt m \lt 1$. This gives $|\frac{x_{n+1}}{x_n}| \lt m$ for all $n \ge N$
$x_{N+1} \lt mx_N$$x_{N+2} \lt mx_{N+1}\lt m^2x_N$ So for all $n \ge N+1$, $x_n \lt mx_{n-1} \lt m^{n-N}x_N$.
 We are therefore left with $0 \lt x_n \lt Am^n$ where $A = \frac{x_N}{m^n}$. As $\lim_{n\to \infty} Am^n = 0$ as $m\lt 1$,using the Squeeze Theorem, we are able to prove the theorem.

You see, the whole thing is dependent on one assumption that $l+\epsilon \lt 1$. But this should ideally hold true for any $\epsilon$. I wouldn't call this proof 'complete'!
Here's another such proof of the quotient law for limits:

Let $\epsilon, k \gt 0.$ Then $\frac{\epsilon}{k}$ is also an arbitrary positive number. If $\{x_n\}$ and $\{y_n\}$ are two sequences, we need to prove that the limit of the quotient of the terms equals the quotient of the limits of the terms( say $l$ and $m$).
For a certain $N$, $|\frac{x_n}{y_n} - \frac{l}{m}| = |\frac{m(x_n-l) + l(m-y_n)}{my_n}| \le |\frac{|m||x_n-l| + |l||m-y_n|}{|m||y_n|}| \lt \frac{\epsilon}{ky_n} + \frac{\epsilon}{ky_n}\frac{|l|}{|m|} = \frac{\epsilon}{k}\frac{|m|+|l|}{|m||y_n|} $$
lim_{n \to \infty} y_n = m$ so $lim_{n \to \infty} |y_n| = |m| $.

Let $ 0 <H<|m|$. Then $ |y_n| > H $ for all $n \ge N_0, N_0 \in \mathbb N$

Choose $N' = max\{N_0, N\}$ so that for all $n \ge N',|\frac{x_n}{y_n} - \frac{l}{m}| < \frac{\epsilon}{k}\frac{|l|+|m|}{|m|H}$
Now choose k such that $ \frac{|l|+|m|}{|m|H} < 1$ so that $|\frac{x_n}{y_n} - \frac{l}{m}| < \epsilon$. Q.E.D.


The last part again contains too convenient choices of constants. I think this might mean that unless you are choosing them in such a manner, the theorem won't hold. It's as though we are creating the proof such that the theorem comes true, which I find strange.
Hopefully I've made myself clear. I wonder if there exist 'more convincing' and more elegant proofs which do not take into account so many arbitrary constants. Thank you!
Edit As suggested in one of the comments, I am inserting a theorem whose proof seems elegant to me-the Squeeze Theorem.

Theorem:Given that $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ are three sequences where $x_n \le y_n \le z_n $ for all $n \ge N,$ where $N \in \mathbb N$, and
$\lim_{n\to\infty } x_n = \lim_{n\to\infty } z_n = l,$ then $lim_{n\to\infty } y_n = l$
Proof: For a given $\epsilon \gt 0$, we have natural numbers $N_1$ and $N_2$ such that $|x_n-l| < \epsilon$ for all $n \ge N_1$ and $|z_n-l| < \epsilon$ for all $n \ge N_2$.
Let $N_3 = max\{N_1, N_2\}$, then for all $n \ge N_3$, $|x_n-l| < \epsilon$ and $|z_n-l| < \epsilon$. 
This means $l-\epsilon < x_n<l+\epsilon$ and $l-\epsilon < z_n<l+\epsilon$ for all $n \ge N_3$. Let $N_4 = max\{N, N_3\}$. Then it holds that $l-\epsilon < x_n < y_n < z_n <l+\epsilon$ and therefore $l-\epsilon < y_n<l+\epsilon$ or  $|y_n-l| < \epsilon$. Q.E.D

We certainly have considered multiple constants here, but we are not arbitrarily assigning them values/choosing them to satisfy certain equations, like so: '$l+\epsilon<1$' or 'choose k such that $ \frac{|l|+|m|}{|m|H} < 1$ '.
 A: These kinds of proofs rely on careful analysis of what you want to prove. Usually, you want to prove something of the form:
For every number $A$ there exists a number $B$ with certain properties related to $A$.
To prove such a statement, you take the number $A$ as a given because the theorem requires the statement to hold for all $A$. So no convenient choices allowed here. But then your task is to find one specific possible choice for $B$, because the statement is only that at least one such number exists.
For this reason it is often perfectly fine to make convenient choices in order to construct one specific convenient choice for $B$.
A: lets address the first proof...
There are two epsilons hiding in this proof.
One for the proposition we seek to prove $\forall \epsilon>0,  \exists N \in \mathbb{N}$ such that $n>N\implies |x_n|<\epsilon$
We cannot choose the value of this epsilon.
But there is a second epsilon...
$\forall \epsilon>0,  \exists N \in \mathbb{N}$ such that $n>N\implies \left||\frac {x_{n+1}}{x_n}| - l\right|<\epsilon$
This comes because it is given that this limit exists, and it is true for all values of $\epsilon.$  This means that we can choose the value for epsilon here and apply the results it generates to evaluate the other limit.
A: $\epsilon$ is a small number, not "any"
number. Of course adding more words will help. The proof you posted assume that the reader has many background knowledge. For example:

Theorem: Let $\{x_n\}$ be a sequence of $\mathbb R^+$ such that $\lim_{n\to\infty} |\frac{x_n+1}{x_n}| = l$. If $0\le l \lt 1$, $\lim_{n\to\infty } x_n = 0$.
Proof: Consider $\epsilon \gt 0$ such that $l+\epsilon \lt 1$. There exists $N \in \mathbb N$ such that $|\frac{x_n+1}{x_n}| \lt \epsilon$ for all $n \ge N$.

Here, "$\epsilon \gt 0$ such that $l+\epsilon \lt 1$" means that for any positive $\epsilon$ small enough that $l + \epsilon \lt 1$, the conclusion holds. We can always find at least one such $\epsilon > 0$ since $l < 1$ and so picking $0 < \epsilon = \frac{1 - l}{2}$ works.
I hope it helps!

Reply to the comments:
The logic go like this:
We want to prove that there exists $\epsilon$ such that $l+\epsilon<1$
To prove the existence, one example suffice. One example can be $\epsilon=\frac{1-l}{2}$.
In fact, for any $\epsilon<1-l$, this will also work.
You cannot arbitrarily choose $\epsilon$ from the whole real line in this example.
DougM's "choose" means "arbitrarily choose"; to further clarify, his "choose $\epsilon$" means that $\epsilon$ can be chosen from any positive real numbers.
In his second example the $\epsilon$ can be arbitrarily chosen from any positive numbers. In his first example, the $\epsilon$ can only be chosen from a restricted set of numbers.
A: I think the first proof is just badly worded, presumably because the author (or publisher) wanted it to be shorter.
Instead of…
“consider $\epsilon>0$ with $l+\epsilon<1$“
I would have written ….
“Since $l<1$, we can find $\epsilon$ such that $\epsilon>0$ and $l+\epsilon<1$“
I think it’s clear that such an $\epsilon$ exists. If there’s any doubt about this, the explicit choice $\epsilon= \tfrac12(1-l)$ makes things definite.
The rest of the proof uses the $\epsilon$ that we’ve chosen. But it doesn’t make any assumptions about this $\epsilon$, other than the assumptions $\epsilon >0$ and $l +\epsilon <1$, which we know to be true.
There’s nothing wrong with the logic of the proof, but I’d say that the way it’s worded is not very helpful.
A: From Wofsey's comment I realized why I 'felt' that the original answer was not adequate.
Revised version:
That is a fairly natural question that you can have as a beginner. I think this explanation should help.
$$
P_1)~~ \exists \epsilon_2 > 0: \epsilon_2 + l < 1 \Rightarrow \forall \epsilon > 0 : \exists N \in \mathbb{N}:\forall n \ge N: |x_n| < \epsilon
$$
However, what you want to show is actually
$$
P_2)~~ \forall \epsilon > 0 : \exists N \in \mathbb{N}:\forall n \ge N: |x_n| < \epsilon
$$
Fortunately, it is always guaranteed that $\exists \epsilon_2 > 0: \epsilon_2 + l < 1 $ as exemplified by $\epsilon_2 = (1-l)/2$, which will suffice for $P_2$.
This kind of technique is used quite often in the proofs of propositions. So you have to make yourself familiar with the approach.
