How can we know when the limit given result is wrong if we are trying to prove it by its definition? I am currently learning Real Analysis and we learned about the definition of a convergent limit and did some exercises to apply the definition in order to prove the limit's result (i.e: "Use the definition of a limit to show $\lim_{x\rightarrow\infty}f(x)=L$)
However, my concern is that what if the given result (i.e: $L$) is wrong? It looks to me that I can prove any limit converges to anything this way (I am not worried about this in an exam context or just, that is just some general thoughts).
To examine this, I tried proving a wrong result myself, took this:

Use the definition of a limit to show that
$\lim_{n\rightarrow\infty}\frac{1}{n}=10$

I worked it out just like I did for the real result.
Solution:
Note: $n,n_0\neq 0$
Let $\epsilon>0$ be arbitrary. We search for $n_0 \in \mathbb{N}$ s.t. $n\geq n_o \Longrightarrow |\frac{1}{n}-10|<\epsilon$
$\frac{1-10n_0}{n_0}<\epsilon \iff n_0>\frac{1}{10+\epsilon}$
We can take, $n_0=\left \lfloor{\frac{1}{10+\epsilon}+1}\right \rfloor $
Done.
Obviously, I don't think that I actually proved that this limit converges to $10$, I just think I am missing something in terms of my understanding or the way this was presented to me was wrong which is why I am posting here.
 A: Let's check if this holds: Let us choose $\epsilon = 1$ and according to your derivation $n_0 = 1$, which means it should hold for $n \geq n_0 = 1$. So let's pick $n=1$. If we plug that into the inequality we get
$$\left | \frac{1}{1} - 10 \right| \leq 1$$
which is clearly wrong. So what went wrong here?
In this case you made a mistake when dealing with the absolute value and the inequality - I've seen this mistake a few times as a teacher, and I can assure you that it is definitely not possible to prove that a converging sequence converges to any arbitary limit:)
So when we try to solve that inequality, we get two cases when it comes to resolving the absolute value:
1. Case: $\frac{1}{n} - 10 \geq 0$: Clearly this cannot happen as $n \geq 1$.
2. Case: $\frac{1}{n} - 10 < 0$:
Then we get  $-\left(\frac{1}{n} - 10\right) < \epsilon$ and I think you can take it from here.
A: Per the comment of Clayton:
$\displaystyle \left|\frac{1 - 10n_0}{n_0}\right| < \epsilon \iff -\epsilon < \frac{1 - 10n_0}{n_0} < \epsilon$ 
$\displaystyle \iff
-n_0\epsilon < 1 - 10n_0 < n_0\epsilon.$
Above assertion justified by the presumption that $n_0 > 0$.
