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$ '.
