3
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ 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). $\endgroup$
    – Clayton
    Sep 23, 2021 at 12:59
  • 1
    $\begingroup$ Assuming Real Analysis is consistent, you will not be able to prove any limit converges to any $L$ $\endgroup$ Sep 23, 2021 at 13:00

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ Non Hausdorff says hi. Jokes aside, yea this sort of thing is quite common for first time analysis students. $\endgroup$ Sep 23, 2021 at 13:03
  • $\begingroup$ thanks for this! $\endgroup$
    – Sergio
    Sep 23, 2021 at 13:10
  • $\begingroup$ You're welcome! Just as a tip I like recommending using the literal definition of absolute values for problems like these: $|x| = \begin{cases} x & \text{ if } x \geq 0 \\ -x & \text{ if } x < 0 \end{cases}$. $\endgroup$
    – flawr
    Sep 23, 2021 at 13:12
  • $\begingroup$ By that, do you mean just taking the 2 cases like you did or what exactly? $\endgroup$
    – Sergio
    Sep 23, 2021 at 18:56
  • $\begingroup$ Yep exactly, in my experience most errors happen by not considering the second case correctly! $\endgroup$
    – flawr
    Sep 23, 2021 at 19:00
3
$\begingroup$

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 $n_0 > 0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .