I have to solve a lot of excercises like these but I am not sure of my answers and in some cases I do not what to do.
Give a sequence $X_n$ and a number L such that:
1) $\forall \epsilon >0 $ and $\forall N \in \Bbb N$ such that if $n \ge N$ then $|X_n - L| < \epsilon $
I did: $a_n=1$ then $|a_n-1|=0<\epsilon$
2)$\forall \epsilon >0 $ $\exists N \in \Bbb N$ such that if $n \le N$ then $|X_n - L| < \epsilon $
I did the same: $a_n=0$ then $|a_n-0|=0<\epsilon$
3)$\forall \epsilon >0 $ $\exists N \in \Bbb N$ such that if $N \le n \le N+2$ then $|X_n - L| \ge \epsilon $
I do not have an idea at all.
4)$\exists \epsilon >0 $ such that $\forall N \in \Bbb N$ if $ n \ge N$ then $|X_n - L| \ge \epsilon $
In this case I think it is not possible give a sequence like this because if this happens then $a_n$ is divergent.
5)$\exists \epsilon >0 $ $\exists N \in \Bbb N$ such that if $n \neq N $ then $|X_n - L| = \epsilon $
6) $\forall \epsilon >0 $ $\exists N \in \Bbb N$ such that if $n \ge L$ then $|\epsilon - N| < X_n $
7)$\exists \epsilon >0 $ $\exists N \in \Bbb N$ such that if $n \ge N $ then $|X_n - L| \ge n\epsilon $
8)$\forall \epsilon >0 $ $\exists N \in \Bbb N$ such that if $n \ge \epsilon $ then $|X_n - L| \ge N $
Any ideas or hint?
Case 1 and 2: You want $|a_n - L| < \epsilon$. You showed $|an - 1| < \epsilon$ in case 1 and $|an - 0| < \epsilon$ in case 2 which wasn't asked for.
In both cases let $a_n = L$ and you're good.
Case 3. You need to define N in terms of epsilon. As you want things to get bigger then epsilon you want it to diverge. Any divergent sequence will work. As you will want to subtract L, I suggest $a_n = n + L$. Then for any $\epsilon > 0$, if $N \ge \epsilon$ then for $N \le n \le N+2$, $|a_n - L| = |n + L -L| =n >= N \ge \epsilon$.
Weird but works.
Case 4: It does have to be divergent. But the exercise never said divergence wasn't allowed. But this doesn't actually need to be divergent as $\epsilon$ is not arbitrary in this case. Let epsilon be ... whatever you like. $\epsilon = 27$, say. Then you need to find a sequence, and an N such that if n $\ge$ N then $|a_n - L|$ $\ge$ 27. Such an N could be anything arbitrary so N = 3. Then {$a_n$} = {97, -2, 27+ L, 28 + L, -L - 29, 500 + L, 27 + L, 31 + L......} will do. Or something less absurd.
Okay, this exercise is weird! But I sort of like it. It forces you to think about exactly what is being asked and literally what does it mean. But I don't like that it plays on expectations and completely thwarts them. Usually we use these N, epsilon to show convergence with the expectation that the epsilon is an arbitrarily small value with we try to squeeze values of the sequence between, and N is a corresponding large value for which index value or terms will be squeezed. In this exercise, this values are nothing at all of the kind.
Case 5: Okay, some (not all) epsilon, a value of N and for all n $\ne$ N, $|a_n -L| = \epsilon$. So epsilon is a constant. $a_n - L = \pm \epsilon$ for all n $\ne$ N another constant. Okay... let epsilon be 1. N = 35. $a_n$ = L + 1 for all n $\ne$ to 35. $a_{35} = \frac{-5,726}{\pi}$.
... and so on....