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.

• Double check the line where you have $|\frac1n-10|<\varepsilon$ and go to the line with no absolute values. You should have a $-\varepsilon$ on the other side (which is where the problem will make itself apparent). Sep 23 at 12:59
• Assuming Real Analysis is consistent, you will not be able to prove any limit converges to any $L$ Sep 23 at 13:00

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.

• Non Hausdorff says hi. Jokes aside, yea this sort of thing is quite common for first time analysis students. Sep 23 at 13:03
• thanks for this! Sep 23 at 13:10
• You're welcome! Just as a tip I like recommending using the literal definition of absolute values for problems like these: $|x| = \begin{cases} x & \text{ if } x \geq 0 \\ -x & \text{ if } x < 0 \end{cases}$. Sep 23 at 13:12
• By that, do you mean just taking the 2 cases like you did or what exactly? Sep 23 at 18:56
• Yep exactly, in my experience most errors happen by not considering the second case correctly! Sep 23 at 19:00

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 $$n_0 > 0$$.