How Does Choosing N Affect Limit Proofs in Calculus?

In summary: But you are trying to proof that ##L## is ##0##, so why don't you do it?But you are trying to proof that L is 0, so why don't you do it?I just did, right? I proved that ##\operatorname{lim} \frac{1}{n^2} = 0## in post #6. I think you are saying I should show that ##L## is ##0## by taking ##\frac{1}{n^2} < \epsilon## to show that ##L## is less than any positive real number ##\epsilon##... is that right?Sorry for asking dumb questions or making errors.. I guess I'm just unsure whether I'm
  • #1
fishturtle1
394
82

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?
 
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  • #2
Your treatment of absolute values is incorrect. For all ##n\geq 1##, ##\lvert 1/n^2 - 4\rvert = 4-1/n^2##.
 
  • #3
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|\,?##
 
  • #4
fresh_42 said:
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.
 
  • #5
Orodruin said:
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!)
 
  • #6
Orodruin said:
Your treatment of absolute values is incorrect. For all ##n\geq 1##, ##\lvert 1/n^2 - 4\rvert = 4-1/n^2##.
fresh_42 said:
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
 
  • #7
fishturtle1 said:
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?

fishturtle1 said:
I was only thinking whether ##\frac {1}{n^2}## could be negative.
Think about it -- ##\frac 1 {n^2}## can't possibly be negative.
 
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  • #8
Mark44 said:
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
 
  • #9
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:
fishturtle1 said:
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.
 
Last edited:
  • #10
fresh_42 said:
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##?
 
  • #11
fishturtle1 said:
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.
 
  • #12
fresh_42 said:
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.
 
  • #13
fishturtle1 said:
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##.
 

Related to How Does Choosing N Affect Limit Proofs in Calculus?

What is a limit proof and why is it important?

A limit proof is a mathematical method used to prove the existence of a limit for a given function. It is important because it allows us to understand the behavior of a function as it approaches a certain value and make predictions about its behavior at that point.

What does N represent in a limit proof?

N represents the number of terms in a sequence that is being used to approximate the limit of a function. It is often used as a variable to represent a large but finite number in order to show that the sequence approaches the limit as the number of terms increases.

How do you choose the appropriate value for N in a limit proof?

The value of N is chosen based on the desired level of accuracy for the limit approximation. In general, a larger value of N will result in a more accurate approximation. It is also important to consider the behavior of the function and the specific point at which the limit is being evaluated.

What are some common techniques used in limit proofs?

Some common techniques used in limit proofs include the squeeze theorem, the epsilon-delta definition of a limit, and the use of L'Hopital's rule. These methods can help simplify complex limit problems and provide a more concrete understanding of the behavior of a function.

What are some tips for successfully completing limit proofs?

Some tips for successfully completing limit proofs include carefully reading and understanding the problem, identifying any known properties or theorems that may be applicable, and breaking the problem into smaller, more manageable steps. It is also important to check your work and make sure all steps are clearly and logically explained.

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