Proving the limit of a sequence from the definition of limit

[marksyncm]

Say that we are asked to prove, using the definition of limits, that the sequence $\frac{4n^2+3}{n^2+n+2}$ tends to $4$ as $n$ tends to infinity. The following is a screenshot of the solution I found in a YouTube video:

(Note that in the definition above, "g" denotes the limit - in this case $4$).

I understand (or at least I think I do) what is happening up to the point I've marked in red - basically it appears to me that we are trying to answer the question: "for which $n$ will the distance between $a_n$ and $4$ be smaller than $\epsilon$? This is why we are solving the absolute value inequality.

So we simplify the absolute value to $\frac{4n+5}{n^2+n+2}$. However, I do not understand why we are saying that this is smaller than $\frac{4n+5n}{n^2}$ or why we chose $\frac{4n+5n}{n^2}$ specifically. What's the purpose behind this? Is it to try and have only one $n$ in the fraction?

For example, if we were attempting to show that $\lim_{n \to \infty} \frac{1}{n} = 0$, we would start from $|\frac{1}{n} - 0| < \epsilon$ meaning that $\frac{1}{n} < \epsilon$ meaning that $n > \frac{1}{\epsilon}$. This means that you give me $\epsilon > 0$, and I'll give you a number $n_0 = \frac{1}{\epsilon}$, and for all $n > n_0$ you will find that $a_n$ is within a distance of less than $\epsilon$ from $0$, which concludes the proof (please correct me if I'm wrong here).

My question is, why can't we just do the same in the example from the screenshot above? What's the purpose of the additional step of saying that the fraction is smaller than <some intermediate value>, and why choose $\frac{4n+5n}{n^2}$ specifically as that value?

Thanks.

 

[FactChecker]

That step increases the numerator and decreases the denominator. Both result in a larger number. It is done to have a simpler expression in $n$, so that $n > n_0$ gives the desired conclusion.

 

[marksyncm]

Isn't the new term completely different from the original sequence, though? We say that $\frac{4n+5}{n^2+n+2} < \frac{4n+5n}{n^2} < \epsilon$, but couldn't we just say, for instance, $\frac{4n+5}{n^2+n+2} < 1000n < \epsilon$ which would give us that $n < \frac{\epsilon}{1000}$ (which doesn't make sense, it seems)?

 

[Mark44]

Isn't the new term completely different from the original sequence, though? We say that $\frac{4n+5}{n^2+n+2} < \frac{4n+5n}{n^2} < \epsilon$, but couldn't we just say, for instance, $\frac{4n+5}{n^2+n+2} < 1000n < \epsilon$ which would give us that $n < \frac{\epsilon}{1000}$ (which doesn't make sense, it seems)?

Right, that doesn't make much sense. For one thing, it's highly unlikely that $1000n$ will be smaller than some arbitrary $\epsilon$.

You're trying to find some number N such that, if n > N, $\frac{4n + 5}{n^2 + n + 2} < \epsilon$. You want to show that the rational expression on the left gets "small" as n gets large.

 

[FactChecker]

The important thing is that $\frac{4n+5n}{n^2}$ is simpler and larger than the original, yet becomes as small as desired when $n \rightarrow \infty$. Therefore, for any $\epsilon > 0$, you can calculate $n_0$ where $n > n_0$ makes $\frac{4n+5n}{n^2} < \epsilon$. That is clearly not true for your proposed $1000n$.

 

[marksyncm]

Thank you. So the new term needs to be larger than the original while also converging to zero (or whatever limit we are working with) as $n$ approaches $\infty$? And, of course, it needs to be in such terms of $n$ that allow us to obtain a value for $n_0$ in terms of $\epsilon$ only?

Also, regarding the screenshot above: the conclusion is that $n_0 = \frac{9}{\epsilon} + 1$. Where does the $+1$ come from? Is it to account for the fact that $n_0$ must be equal to 1 or larger (as it is a natural number), so we add the $+1$ to make sure this is the case for very large $\epsilon$ values?

 

[Mark44]

Thank you. So the new term needs to be larger than the original while also converging to zero (or whatever limit we are working with) as $n$ approaches $\infty$? And, of course, it needs to be in such terms of $n$ that allow us to obtain a value for $n_0$ in terms of $\epsilon$ only?

Yes to both.

Also, regarding the screenshot above: the conclusion is that $n_0 = \frac{9}{\epsilon} + 1$. Where does the $+1$ come from? Is it to account for the fact that $n_0$ must be equal to 1 or larger (as it is a natural number), so we add the $+1$ to make sure this is the case for very large $\epsilon$ values?

Just before that, they have $\frac 9 \epsilon < n$. Multiply both sides by $\epsilon$ and n. Both are positive, so the direction of the inequality doesn't change.

BTW, the writing of whoever did the work in the screenshot is very unclear. The 9's are almost unrecognizable. If the previous work didn't have 4 + 5, I wouldn't have known that that squiqqle was a 9.

 

[FactChecker]

Thank you. So the new term needs to be larger than the original while also converging to zero (or whatever limit we are working with)

It must be zero, no other limit is appropriate.

as $n$ approaches $\infty$? And, of course, it needs to be in such terms of $n$ that allow us to obtain a value for $n_0$ in terms of $\epsilon$ only?

There might be other parameters involved. The selection of $n_0$ might depend on those parameters. But not in this case (or most cases).

Also, regarding the screenshot above: the conclusion is that $n_0 = \frac{9}{\epsilon} + 1$. Where does the $+1$ come from? Is it to account for the fact that $n_0$ must be equal to 1 or larger (as it is a natural number), so we add the $+1$ to make sure this is the case for very large $\epsilon$ values?

Actually, $\frac{9}{\epsilon}$ is probably not an integer since $\epsilon$ is an arbitrary positive real, and adding 1 does not make it an integer. But that is ok. We could define the real number $r_0 = \frac{9}{\epsilon} + 1$ and use it the same way. Adding +1 was probably done just to make the proof a little safer (just in case his value of $n_0$ gave equality rather than $>$). He could have added any large natural number, since he only needed to prove something for large $n$.

CORRECTION: I overlooked the brackets, which take the integer part: $[\frac{9}{\epsilon}]$. So this is an integer and the '+1' makes sure that $n_0$ is large enough after the fractional part of $\frac{9}{\epsilon}$ is removed.

 

[PeroK]

Actually, $\frac{9}{\epsilon}$ is probably not an integer since $\epsilon$ is an arbitrary positive real, and adding 1 does not make it an integer. But that is ok.

The solution above uses $[\frac{9}{\epsilon}]$, which I assume is the integer part of $\frac{9}{\epsilon}$, and then adds $1$ to it.

@marksyncm adding $1$ is to ensure that $n_0 > \frac{9}{\epsilon}$.

 

[PeroK]

Isn't the new term completely different from the original sequence, though? We say that $\frac{4n+5}{n^2+n+2} < \frac{4n+5n}{n^2} < \epsilon$,

This is such an important and powerful idea (and actually quite simple) that you really want to make sure you've digested it. For example, suppose you want to prove that the sequence $2n^3 + 6n^2 + 3n -1$ diverges.

Your first strategy might be to try to find the point $n_0$ at which this sequence exceeds a given number. So, you would look for a solution to $2n^3 + 6n^2 + 3n -1 = M$, where $M$ is some (large) number. But, it not easy to solve a cubic equation. So that looks a hard way to do it.

Alternatively, you could look for another much simpler sequence that is less than the one you have. There are then lots of ways of doing this.

E.g. you could note that $\forall n: \ 3n - 1 > 0$ and $6n^2 > 0$. That means that:

$\forall n: 2n^3 + 6n^2 + 3n -1 > 2n^3$

And, if you can find $n_0$ for the sequence $2n^3$, then that $n_0$ does for the original sequence as well.

However, you could go further and note that: $\forall n: 2n^3 > n$

And now you have $\forall n: 2n^3 + 6n^2 + 3n -1 > n$.

That means that $\forall M$ you can simply take $n_0 = M$ or, if you are being pedantic, $n_0 = [M] + 1$ and:

$\forall M > 0, \ n > [M]+1 \ \Rightarrow 2n^3 + 6n^2 + 3n -1 > M$

And that proves that the sequence diverges.

In any case, this general idea of "crude" approximations is very powerful and can make things much simpler.

 

[marksyncm]

Thank you all, this was very helpful.

Question regarding this:

(...)
And, if you can find $n_0$ for the sequence $2n^3$, then that $n_0$ does for the original sequence as well.
.

Thanks for the explanation, this makes a lot of sense. What happens if we can't find an $n_0$ for this one, though (not in this example, of course)? Because the new sequence is smaller than the original one, couldn't it "behave" differently than the original one?

 

[PeroK]

Thank you all, this was very helpful.

Question regarding this:Thanks for the explanation, this makes a lot of sense. What happens if we can't find an $n_0$ for this one, though (not in this example, of course)? Because the new sequence is smaller than the original one, couldn't it "behave" differently than the original one?

Yes, obviously, you need to make sure your new sequence still has the required property. For polynomials in general the leading term (i.e. the highest power) dominates.

 

[marksyncm]

To make sure I understood, here's another example that I'm going to try and solve: $\lim{n \to \infty} (\frac{n^3}{2n^+9} - \frac{n}{2}) = 0$. So we have that:

$$|\frac{n^3}{2n^2+9} - \frac{n}{2} - 0| = |\frac{2n^3-2n^3-9n}{4n^2+18}| = \frac{9n}{4n^2+18} < \epsilon$$

Now I could say that

$$\frac{9n}{4n^2+18} < \frac{9n}{4n^2} =\frac{9}{4n} < \epsilon \rightarrow n > \frac{9}{4\epsilon}$$

So we can say that $n_0 = \frac{9}{4\epsilon} + 1$.

Sounds good?

 

[marksyncm]

Yes, obviously, you need to make sure your new sequence still has the required property. For polynomials in general the leading term (i.e. the highest power) dominates.

Got it, thanks!

 

[PeroK]

To make sure I understood, here's another example that I'm going to try and solve: $\lim{n \to \infty} (\frac{n^3}{2n^+9} - \frac{n}{2}) = 0$. So we have that:

$$|\frac{n^3}{2n^2+9} - \frac{n}{2} - 0| = |\frac{2n^3-2n^3-9n}{4n^2+18}| = \frac{9n}{4n^2+18} < \epsilon$$

Now I could say that

$$\frac{9n}{4n^2+18} < \frac{9n}{4n^2} =\frac{9}{4n} < \epsilon \rightarrow n > \frac{9}{4\epsilon}$$

So we can say that $n_0 = \frac{9}{4\epsilon} + 1$.

Sounds good?

Yes, you've got the idea.

 

[marksyncm]

Thanks again, everyone.

 

[FactChecker]

The solution above uses $[\frac{9}{\epsilon}]$, which I assume is the integer part of $\frac{9}{\epsilon}$, and then adds $1$ to it.

@marksyncm adding $1$ is to ensure that $n_0 > \frac{9}{\epsilon}$.

Oh. I overlooked that symbology. Good catch. I stand corrected.