I need to understand those things better. Say I have to check for the correctness, or not, of the limit $$\lim_{n \to +\infty} \frac{1}{n+1} = 1$$ Now I know the limit is false, but I want to verify with the definition, closing with a contradiction or a violation of the same.
So I know that in this case $\forall \epsilon > 0, \exists M_{\epsilon} > 0$ such that when $n > M$ we have $| f(n) - \ell | < \epsilon$.
Then here $$\Big| \frac{1}{n+1} - 1\Big| < \epsilon \implies \Big| \frac{-n}{n+1}\Big| < \epsilon $$
Hence we can solve getting $n > \frac{-\epsilon}{\epsilon -1}$.
Then it suffices to take $M_{\epsilon} = \frac{-\epsilon}{\epsilon -1}$
Can we say the limit is false because this does violate the statement? I mean the part $\forall \epsilon > 0, \exists M > 0$. This is false if we choose $\epsilon = 1$ or $\epsilon = 1.01$.
In regart to my previous point: is it legit to choose such a "large" $\epsilon$? I mean I know the definition reads $\forall \epsilon > 0$, but I sometimes (wrongly) think that we "need" always something like $\epsilon \to 0$, I mean positive but very little.
I wanted to ask another question here, but I'll ask it in a different topic.
Thank you for the moment.
I am not sure about your question, but actually you can use the negative proposition for help.
The definition of limit of a sequence is that $$\forall \varepsilon > 0 \exists N \in \mathbb N \forall n > N(|a_n - a| < \varepsilon).$$ Therefore the negative proposition of it should be $$\exists \varepsilon > 0 \forall N \in \mathbb N \exists n > N(|a_n - a| \geq \varepsilon).$$ What you need to prove is that $$\lim_{n \rightarrow \infty}\frac 1 {n + 1} \neq 1.$$ So the goal is to find such an $\varepsilon$. Then here we have $$|\frac 1 {n + 1} - 1| = |\frac n {n+1}| \geq \varepsilon,$$ if we take an arbitrary $\varepsilon$, like $\varepsilon = 1 / 2$ just as @Surb said, then that will be $n \geq 1$, so we say that we have found an $\varepsilon = 1 / 2$ that for all $N \in \mathbb N$, we can choose $n = \max\{1, N + 1\} > N$ s.t. $\left|\dfrac 1 {n + 1} - 1\right| \geq \varepsilon = \dfrac 1 2$, which proves what you want.
In your question you say that you are not sure whether choosing a relatively big number is legit, but actually you can verify that if one $\varepsilon$ don not satisfy the positive proposition, all $\varepsilon$ smaller than it do not so according to the negetive proposition.
You can imagine an interval of $(1 - \varepsilon, 1 + \varepsilon)$ and let $\varepsilon$ change from a big number to small one. When it is big enough, it can contain all infinite terms of $a_n$. But at some points it fails, like the example of $1/2$. So only finite terms will fall into the interval, which is contradictory to the definition, which says all $x_n (n > N)$ fall into a neighbourhood of $a$, indicating that there are infinite terms in it.