How Does Choosing N Affect Limit Proofs in Calculus?

[fishturtle1]

Homework Statement

Find and prove $\operatorname{lim} \frac {1}{n^2}$.

Homework Equations

In the textbook, we assume that the limit is going to infinity without writing it.

If L is the limit, we have for all $\epsilon > 0$, there exists $N$ such that $n \epsilon \mathbb{Z}$ and $n > N$ implies $\vert \frac{1}{n^2} - L \vert < \epsilon$.

The Attempt at a Solution

This is an example in the book. I can follow how to solve for N if we guess $\operatorname{lim} \frac {1}{n^2} = 0$.
We just have $\vert \frac {1}{n^2} - 0 \vert < \frac {1}{n^2} < \epsilon$.
So $n > \frac {1}{\sqrt{\epsilon}}$. Then we work backwards in the actual proof.

I am confused, if we guess $\operatorname{lim} \frac {1}{n^2} = 4$ where 4 is just some arbitrary number... then we have,
$\vert \frac {1}{n^2} - 4 \vert = \frac {1}{n^2} - 4 < \frac {1}{n^2} < \epsilon$. So, $n^2 > \frac {1}{\epsilon}$. So $n > \frac {1}{\sqrt{\epsilon}}$.

So..
Proof: Let $\epsilon > 0$ and choose $N = \frac {1}{\sqrt{\epsilon}}$. Suppose $n > N$ and $n$ is an integer. Then $n > \frac {1}{\sqrt{\epsilon}}$. This implies $n^2 > \frac {1}{\epsilon}$ which implies $\frac {1}{n^2} < \epsilon$. So $\epsilon > \frac {1}{n^2} > \frac {1}{n^2} - 4 = \vert \frac {1}{n^2} - 4 \vert$. This proves that $\operatorname{lim} \frac{1}{n^2} = 4.$

But limits are unique... but I'm not sure what I'm doing wrong or why this doesn't work?

 

[Orodruin]

Your treatment of absolute values is incorrect. For all $n\geq 1$, $\lvert 1/n^2 - 4\rvert = 4-1/n^2$.

 

[fresh_42]

Beside that the limit is obviously not at infinity, at least not for increasing $n$, which you failed to mention, and that $\frac{1}{n^2} \nless \frac{1}{n^2}$, how comes that $\frac{1}{n^2}-4=|\frac{1}{n^2}-4|\,?$

 

[Orodruin]

Beside that the limit is obviously not at infinity, at least not for increasing n

I think he means $n \to \infty$, not that the limit is infinite.

 

[fresh_42]

I think he means $n \to \infty$, not that the limit is infinite.

When playing blitz at university, one of the standard complaints had been: "Sauber setzen!" (set properly; exakt set!)

 

[fishturtle1]

Your treatment of absolute values is incorrect. For all $n\geq 1$, $\lvert 1/n^2 - 4\rvert = 4-1/n^2$.

Beside that the limit is obviously not at infinity, at least not for increasing $n$, which you failed to mention, and that $\frac{1}{n^2} \nless \frac{1}{n^2}$, how comes that $\frac{1}{n^2}-4=|\frac{1}{n^2}-4|\,?$

Thank you for the replies, I said $\frac {1}{n^2} - 4 = \vert \frac {1}{n^2} - 4 \vert$ because I was only thinking whether $\frac {1}{n^2}$ could be negative. I didn't think about if/when $\frac {1}{n^2} - 4$ could be negative.
I see now that for any nonnegative integer $n$, $\frac {1}{n^2} - 4 < 0$ and so $\frac {1}{n^2} - 4 \neq \vert \frac {1}{n^2} - 4 \vert$. So it must be that $\vert \frac {1}{n^2} - 4 \vert = 4 - \frac {1}{n^2}$.

Also, in my original post I made a mistake, I meant to say I'm find $\operatorname{lim}\frac {1}{n^2}$ as $n \rightarrow \infty$, not that the limit is going to infinity.

So now I've tried to "fix" my algebra to get some other limit besides 0 but i run into dead ends.
$\vert \frac {1}{n^2} - 4 \vert = 4 - \frac {1}{n^2} < \epsilon$. So $- \frac {1}{n^2} < \epsilon - 4$. So $n^2 < \frac {1}{\epsilon - 4}$. So $n < \frac{1}{\sqrt{\epsilon - 4}}$. But how do we show that $\operatorname{lim} \frac{1}{n^2} < 0$? Is there a direct way to do this, or do we do that by showing $\operatorname{lim} \frac{1}{n^2} = 0$ and then taking advantage of the theorem that limits are unique?

Then I thought maybe we could do $\operatorname{lim} \frac {1}{n^2} = -4$.
So, $\vert \frac {1}{n^2} - (-4) \vert = \frac {1}{n^2} + 4 < \epsilon$. So $\frac{1}{n^2} < \epsilon - 4$. So $n^2 > \frac {1}{\epsilon - 4}$. So $n > \frac {1}{\sqrt{\epsilon - 4}}$.. The only problem I see with this is that $\epsilon > 4$.. So that means we can't use this since our statement is "For all $\epsilon > 0$... So if we replaced -4 with -L, would that be sufficient to show $\operatorname{lim} \frac{1}{n^2} \ge 0$?
Edit: yes i think it does

 

[Mark44]

Find and prove $\operatorname{lim} \frac {1}{n^2}$.

I hope it's pretty obvious that $\lim_{n \to \infty} \frac 1 {n^2} = 0$

Instead of messing with numbers such as 4 that couldn't possibly be the limit, why not use your argument to show that $\frac 1 {n^2}$ can be made arbitrarily close to 0?

I was only thinking whether $\frac {1}{n^2}$ could be negative.

Think about it -- $\frac 1 {n^2}$ can't possibly be negative.

 

[fishturtle1]

I hope it's pretty obvious that $\lim_{n \to \infty} \frac 1 {n^2} = 0$

Instead of messing with numbers such as 4 that couldn't possibly be the limit, why not use your argument to show that $\frac 1 {n^2}$ can be made arbitrarily close to 0?

Think about it -- $\frac 1 {n^2}$ can't possibly be negative.

Thank you for the reply. It does seem clear that $\operatorname{lim} \frac{1}{n^2} = 0$ because $n^2$ gets really big, so $\frac{1}{n^2}$ should get really small. I guess I was unsure..and maybe overthinking.. is what would happen if we tried to use the limit definition to prove something else was the limit, like how would that fail?

But I can prove that $\operatorname{lim} \frac{1}{n^2} = 0$.
To solve for N, I have: $\vert \frac{1}{n^2} \vert = \frac {1}{n^2} < \epsilon$, so $n^2 > \frac{1}{\epsilon}$ so $n > \frac{1}{\sqrt{\epsilon}}$.

Proof: Let $\epsilon > 0$ and choose $N = \frac{1}{\sqrt{\epsilon}}$. Suppose $n > N$ and $n$ is a nonnegative integer. Then $n > \frac{1}{\sqrt{\epsilon}}$ implies $n^2 > \frac{1}{\epsilon}$ implies $\frac{1}{n^2} < \epsilon$ implies $\vert \frac{1}{n^2} \vert < \epsilon$. This proves that $\operatorname{lim} \frac{1}{n^2} = 0$. []

I see that $\frac{1}{n^2} > 0$ for all positive integers n.. but I wasn't careful when I assumed that since $\frac{1}{n^2} > 0$ then $\frac{1}{n^2} - 4 > 0$ for all positive integer n.. which I now see is not true

 

[fresh_42]

You still have a sign error in post #6, because it is $n^2 < \frac{1}{4-\varepsilon}$. Anyway, the entire structure isn't optimal. The sequence is never negative, so how should the limit be? You have already said it:

If $L$ is the limit, we have for all $\epsilon > 0$, there exists $N$ such that $n \in \mathbb{Z}$ and $n > N$ implies $\vert \frac{1}{n^2} - L \vert < \epsilon.$

So what to do? We have an arbitrary epsilon given, a guessed limit $L$, here we have $L=0$, and we need to name an $N=N(\varepsilon)$ such that for all greater $n$ the distance of sequence elements from the limit is smaller as our given range $\varepsilon$. So we have to set $\mathbf{N}=N(\varepsilon)$ beforehand. The structure of the proof therefore goes:

Let $\varepsilon > 0$ be arbitrary and set $N=N(\varepsilon) := \ldots$ Now show
$$
\forall n> N(\varepsilon)\, : \, | \dfrac{1}{n^2}-0| < \ldots < \varepsilon \quad (1)
$$

This is the final result. To find $L=0$, we have to guess, which is done by looking at the sequence: $1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25}\ldots$ which makes it plausible, that $L=0$. Once we know that, we can use it in other cases where $\frac{1}{n^2}$ occurs. The more tools and examples we gathered, the more limits we can guess.

The other part is to find $N=N(\varepsilon)$ which I like to write as $N(\varepsilon)$ or $N_\varepsilon$ because it depends on
$\varepsilon$ and isn't a fixed number $N$. To me this is a source of mistakes. If it depends on a variable, then it should be noted. A general rule, which can help to avoid a lot of mistakes. Anyway. Now in order to find it, we write the inequation we are looking for:
$$| \dfrac{1}{n^2}-0| = \dfrac{1}{n^2} < \dfrac{1}{N_\varepsilon^2} < \dfrac{1}{N_\varepsilon} \leq \varepsilon$$
This means, that $N_\varepsilon \geq \dfrac{1}{\varepsilon}$ and $N_\varepsilon < N_\varepsilon^2$. Since $N_\varepsilon$ has to be a natural number, the first condition is achieved be the ceiling function: $N_\varepsilon = \lceil \dfrac{1}{\varepsilon} \rceil$ and the second by the requirement, that $N_\varepsilon > 1$. Thus we set $N_\varepsilon := \operatorname{max}\{2,\lceil \dfrac{1}{\varepsilon} \rceil\}$. This way, we have deduced the requirements for $N_\varepsilon$ and we can write down $(1)$.

Note that the ceiling function only allows a less or equal, but as we had some proper less before, we can guarantee a proper less for $|a_n-L|< \varepsilon$.

This is basically the full solution, because I used this simple example to demonstrate how those proofs are structured and how the logic goes. I really suggest, that you go through this to comprehend the logic behind. Especially the details with the maximum, and the ceiling function often occur in such proofs. I bet you would have forgotten, that the inequality above ($N < N^2$) is wrong for $N=1$. We could have done the proof with less or equal, too, because we already had a proper less in our line, but I wanted to demonstrate, how easy it is to overlook such details and even more, that there is no need to be as close as possible with your $N_\varepsilon$. Make it as big as you like in order to make things easier, e.g. to get rid of the root, resp. square. There is no prize for the smallest $N_\varepsilon$ possible. Any $N_\varepsilon$ will do, and the bigger it is, the faster a proof can be read.

 

[fishturtle1]

You still have a sign error in post #6, because it is $n^2 < \frac{1}{4-\varepsilon}$. Anyway, the entire structure isn't optimal. The sequence is never negative, so how should the limit be? You have already said it:

So what to do? We have an arbitrary epsilon given, a guessed limit $L$, here we have $L=0$, and we need to name an $N=N(\varepsilon)$ such that for all greater $n$ the distance of sequence elements from the limit is smaller as our given range $\varepsilon$. So we have to set $\mathbf{N}=N(\varepsilon)$ beforehand. The structure of the proof therefore goes:

Let $\varepsilon > 0$ be arbitrary and set $N=N(\varepsilon) := \ldots$ Now show
$$
\forall n> N(\varepsilon)\, : \, | \dfrac{1}{n^2}-0| < \ldots < \varepsilon \quad (1)
$$

This is the final result. To find $L=0$, we have to guess, which is done by looking at the sequence: $1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25}\ldots$ which makes it plausible, that $L=0$. Once we know that, we can use it in other cases where $\frac{1}{n^2}$ occurs. The more tools and examples we gathered, the more limits we can guess.

The other part is to find $N=N(\varepsilon)$ which I like to write as $N(\varepsilon)$ or $N_\varepsilon$ because it depends on
$\varepsilon$ and isn't a fixed number $N$. To me this is a source of mistakes. If it depends on a variable, then it should be noted. A general rule, which can help to avoid a lot of mistakes. Anyway. Now in order to find it, we write the inequation we are looking for:
$$| \dfrac{1}{n^2}-0| = \dfrac{1}{n^2} < \dfrac{1}{N_\varepsilon^2} < \dfrac{1}{N_\varepsilon} \leq \varepsilon$$
This means, that $N_\varepsilon \geq \dfrac{1}{\varepsilon}$ and $N_\varepsilon < N_\varepsilon^2$. Since $N_\varepsilon$ has to be a natural number, the first condition is achieved be the ceiling function: $N_\varepsilon = \lceil \dfrac{1}{\varepsilon} \rceil$ and the second by the requirement, that $N_\varepsilon > 1$. Thus we set $N_\varepsilon := \operatorname{max}\{2,\lceil \dfrac{1}{\varepsilon} \rceil\}$. This way, we have deduced the requirements for $N_\varepsilon$ and we can write down $(1)$.

Note that the ceiling function only allows a less or equal, but as we had some proper less before, we can guarantee a proper less for $|a_n-L|< \varepsilon$.

This is basically the full solution, because I used this simple example to demonstrate how those proofs are structured and how the logic goes. I really suggest, that you go through this to comprehend the logic behind. Especially the details with the maximum, and the ceiling function often occur in such proofs. I bet you would have forgotten, that the inequality above ($N < N^2$) is wrong for $N=1$. We could have done the proof with less or equal, too, because we already had a proper less in our line, but I wanted to demonstrate, how easy it is to overlook such details and even more, that there is no need to be as close as possible with your $N_\varepsilon$. Make it as big as you like in order to make things easier, e.g. to get rid of the root, resp. square. There is no prize for the smallest $N_\varepsilon$ possible. Any $N_\varepsilon$ will do, and the bigger it is, the faster a proof can be read.

Thank you for the detailed outline.. I think I can follow most of it and will keep the things you mentioned in mind, like choosing bigger $N(\varepsilon)$ to make the proof more readable/shorter..

I don't understand why we need to have $N_{\varepsilon}$ as a natural number. Doesn't it make sense to just let $N_{\varepsilon} > 0$?

 

[fresh_42]

Thank you for the detailed outline.. I think I can follow most of it and will keep the things you mentioned in mind, like choosing bigger $N(\varepsilon)$ to make the proof more readable/shorter..

I don't understand why we need to have $N_{\varepsilon}$ as a natural number. Doesn't it make sense to just let $N_{\varepsilon} > 0$?

Yes, it's a bit of a convention. Since it measures a certain point in the sequence "for all indices $n$ greater than $N_\varepsilon$ we have for $a_n \ldots$" it makes kind of sense to choose a natural number, because there is no sequence element $a_{4.87632932807924}$ The ceiling function also gives a bit of additional space for the inequalities. Otherwise we would eventually have to bother whether for $\frac{3}{2}$ we have to choose $1$ or $2$. With numbers as in my example here, it is clear, but if the boundary is a complex algebraic expression, things are less obvious.

 

[fishturtle1]

Yes, it's a bit of a convention. Since it measures a certain point in the sequence "for all indices $n$ greater than $N_\varepsilon$ we have for $a_n \ldots$" it makes kind of sense to choose a natural number, because there is no sequence element $a_{4.87632932807924}$ The ceiling function also gives a bit of additional space for the inequalities. Otherwise we would eventually have to bother whether for $\frac{3}{2}$ we have to choose $1$ or $2$. With numbers as in my example here, it is clear, but if the boundary is a complex algebraic expression, things are less obvious.

So for example in post #9, we could have let $N_\varepsilon$ just be nonnegative and this would technically lead to a correct answer, but in the context of the problem, since $n > N_\varepsilon$ and $n$ are just indices(integers) it makes sense to add the condition that $N_\varepsilon$ is a natural number. So its not necessary but its good/standard practice?

Also doesn't using the ceiling take away space from the inequalities? Since $N_\varepsilon > \lceil\frac{1}{\varepsilon}\rceil \ge \frac{1}{\varepsilon}$? Since $N_\varepsilon$ isn't given a looser bound.

 

[fresh_42]

So for example in post #9, we could have let $N_\varepsilon$ just be nonnegative and this would technically lead to a correct answer, ...

What do you mean? $N_\varepsilon$ depends on the choice of $\varepsilon$, so it cannot be chosen fixed. For $N_\varepsilon < N_\varepsilon^2$ we also need greater than $1$, so non-negative isn't sufficient.

... but in the context of the problem, since $n > N_\varepsilon$ and $n$ are just indices(integers) it makes sense to add the condition that $N_\varepsilon$ is a natural number. So its not necessary but its good/standard practice?

Yes, it's just a standard as it marks a natural number, but you are right, it isn't necessary. The next natural number is $n>N_\varepsilon$ anyway. E.g. I'm used to the condition $n \geq N_\varepsilon$ and then it is better to make sure to have the next integer.

Also doesn't using the ceiling take away space from the inequalities? Since $N_\varepsilon > \lceil\frac{1}{\varepsilon}\rceil \ge \frac{1}{\varepsilon}$? Since $N_\varepsilon$ isn't given a looser bound.

Remember that $N_\varepsilon$ may be too big, it doesn't matter, but too small is a problem. E.g. let's take
$$
|a_n - L| = |\dfrac{1}{n^2}-0|=\dfrac{1}{n^2} < \varepsilon = \dfrac{5}{21}
$$
then $\dfrac{1}{\varepsilon} = 4.25$ and with $5$ we're on the secure side. Our estimation is then $N_{5/21} = max\{2,5\}=5$ and for all $n > N_{5/21}$ means for all $n= 6,7,8,\ldots$ and $\dfrac{1}{36}$ is certainly less than $\dfrac{5}{21}$ without any calculations to be made. Be generous, it doesn't matter. There is no prize for searching $N_\varepsilon = N_{5/21}=2$. See, if we had chosen $N_{5/21}=2$ one had to make sure that the condition was $\forall n>2$ and not $\forall n \geq 2$ and whether $\dfrac{1}{4}<\dfrac{5}{21}$ which is wrong, but we had $\forall n>2$ so it starts with $a_3=\dfrac{1}{9}< \dfrac{5}{21}$ which is right. What a waste of time and unnecessary controls. Take $N_{5/21}=1,000$ and everything is apparent at once, without any trouble with decimals. Of course, $N_\varepsilon=1,000$ won't work for $\varepsilon=10^{-9}$, so we have to find another $N_\varepsilon$ in this case, and that's why $N$ depends on $\varepsilon$.