To show a sequence is bounded, monotone and to find its limit

Answers to 1 and 2 are fine.

But you ask a really good question about how do you know what you can accept as obvious or not.

You should have early one spent half a lesson (or told to read) axioms and definitions of the rational/real field.

You therefore have Axiom: if $a < b$ the $a+m < b+m$ for all $m$, and if $c > 0$ than $ac < bc$. And from there you have a proposition that if $1 < k \iff 0< \frac 1k < 1$ (Pf: if $1 < k$ and $\frac 1k \ge 1$ we'd have the contradictory $\frac 1kk \ge 1\cdot k$. If $\frac 1k \le 0$ then $1=\frac 1kk<0\cdot k = 0$. [But note we also have to prove $0\cdot k = 0$, and $1 > 0$ which... well, they should be excercises under your belt]

But once you get through the lesson (which slogs a heck of a lot of stuff; so much stuff that if you actually gave it the time you think it requires you'd never get to lesson 2) you can assume all basic "facts" about numbers.

And .... well, rule of thumb. Feel free to say For all $n > 1$ we have $0 < \frac 1n \le 1$ without justification. BUT be prepared to back it up if you are asked.

Question 3:

Yes you do have to prove that $\lim_{n\to \infty} \frac {n+1}n = 1$. But once you see that $\frac {n+1}n =1+\frac 1n$ that's easy with a $N$ epsilon proof:

$|\frac {n+1}n -1|=|(1+\frac 1n) - 1| = |\frac 1n| = \frac 1n < \epsilon \iff$

$\frac 1n < \epsilon \iff$

$n > \frac 1\epsilon$.

So by the definition of $\lim_{n\to \infty} a_n=L$ we have that there exists an $N: = \frac 1\epsilon$ so that $n > N\implies |\frac {n+1}n -1| < \epsilon$. Thus we have proven $\lim_{n\to \infty} \frac {n+1}n = 1$.

BUT..... again.... once you prove something once you can assume it is known forever.

You have probably already proven 1) If $\lim a_x =L$ then $\lim (a_x + c) = L +c$ for a constant $c$[$*$] and you have probably already proven 2) $\lim_{n\to \infty} \frac 1n =0$[$**$].

If so, you can just state: As $\frac {n+1}n$ is bounded below and is monotonic decreasing the limit exists[$***$] and so $\lim_{n\to \infty} \frac {n+1}n = \lim_{n\to \infty}(1 +\frac 1n) = 1+\lim_{n\to \infty} \frac 1n = 1+0 =1$.

That's it.

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[$*$]As $|(c+ a_x) -(c+L)| = |a_x - L|$ so $|a_x -L| < \epsilon \iff |(c+ a_x) -(c+L)|<\epsilon$ so $\lim a_x = L \iff \lim (c+a_x) = L + c$.

[$**$]ANd $|\frac 1n -0| = \frac n < \epsilon \iff n > \frac 1{\epsilon}$ so $\lim_{n\to \infty}\frac 1n = 0$.

[$***$] Actually this needs to be justified, but presumably you already have. It is the basic property of Real Numbers that if a set is bounded above or below then the $\sup$ or $\inf$ exists. If a set if monotonic and bounded below then. $\lim_{n\to \infty} a_n$ exists and that it must be equal to $\inf a_n$. This is because for any $\epsilon >0$ then $\inf a_n + \epsilon$ is not a lower bound and there a $N$ so that $\inf a_n \le a_N < \inf a_n + \epsilon$. And as $a_n$ is monotonic decrease all $k > N$ are such that $\inf a_n \le a_k < a_N < \inf a_n + \epsilon$. So $|(\inf a_n)-a_k| < \epsilon$. SO $\lim_{n\to \infty} a_n = \inf a_n$.