# Limits

#### Petrus

##### Well-known member
Calculate limit I have none progress, Is there any special rule for this limit?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Calculate limit I have none progress, Is there any special rule for this limit?
You can find a lower and upper bound with an integral.

#### Petrus

##### Well-known member
You can find a lower and upper bound with an integral.
Huh? What you mean?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Calculate limit I have none progress, Is there any special rule for this limit?
You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Here's a picture that is somewhat indicative of how it works. In this example the area of the rectangles is an upper bound of the integral of the function (starting from 1 in this case).
The area of the rectangles is equal to the series of the function.

#### Petrus

##### Well-known member
You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$
So if I got this correct.
I will have
$$\displaystyle \int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk$$ or what shall I integrate to respect?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So if I got this correct.
I will have
$$\displaystyle \int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk$$ or what shall I integrate to respect?
Yep. That's it.

#### Petrus

##### Well-known member
So if I just antiderivate that It Will be $$\displaystyle \frac{3n}{k^3}$$ Is that correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So if I just antiderivate that It Will be $$\displaystyle \frac{3n}{k^3}$$ Is that correct?
Nope.
Consider what the derivative of $$\displaystyle \frac{3n}{k^3}$$ is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$

#### Petrus

##### Well-known member
Nope.
Consider what the derivative of $$\displaystyle \frac{3n}{k^3}$$ is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$
hmm... $$\displaystyle \frac{n}{k+n^2}$$ that is what I get

#### Klaas van Aarsen

##### MHB Seeker
Staff member
hmm... $$\displaystyle \frac{n}{k+n^2}$$ that is what I get
Can you show your steps starting from $$\displaystyle \frac{d}{dk}\left(\frac {3n}{k^3}\right)$$ then?

#### Petrus

##### Well-known member
Can you show your steps starting from $$\displaystyle \frac{d}{dk}\left(\frac {3n}{k^3}\right)$$ then?
I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask $$\displaystyle \frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}$$ so the derivate is $$\displaystyle -9k^{-4} <=> \frac{-9}{k^4}$$

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask $$\displaystyle \frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}$$ so the derivate is $$\displaystyle -9k^{-4} <=> \frac{-9}{k^4}$$
Correct! In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So $$\displaystyle \frac{d}{dk}\left(\frac{3n}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}$$.

So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?

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#### Petrus

##### Well-known member
Correct! In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So $$\displaystyle \frac{d}{dk}\left(\frac{3}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}$$.

So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?
Nop. But when I look in my book it looks like it will be $$\displaystyle \arctan$$

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Nop. But when I look in my book it looks like it will be $$\displaystyle \arctan$$
or Wait I think I got it now!

#### Petrus

##### Well-known member
Nop. But when I look in my book it looks like it will be $$\displaystyle \arctan$$

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or Wait I think I got it now!
$$\displaystyle \frac{3}{x^2+3^2}$$ if we factour out 3 from top and bot we get $$\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}$$ so our integrate become $$\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}$$ Is this correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$\displaystyle \frac{3}{x^2+3^2}$$ if we factour out 3 from top and bot we get $$\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}$$ so our integrate become $$\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}$$ Is this correct?
Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate $$\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}$$?

#### Petrus

##### Well-known member
Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate $$\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}$$?
im confused now haha. 3/3=1 so we got $$\displaystyle \arctan{\frac{x}{3}}$$
edit: if I got this right It is wrong cause if we use chain rule we get $$\displaystyle \frac{1}{x^2+1}\frac{1}{3}$$ and that is not same hmm.. how I integrate that then?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
im confused now haha. 3/3=1 so we got $$\displaystyle \arctan{\frac{x}{3}}$$
edit: if I got this right It is wrong cause if we use chain rule we get $$\displaystyle \frac{1}{x^2+1}\frac{1}{3}$$ and that is not same hmm.. how I integrate that then?
First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).

(Btw, you did correctly find the derivative of $$\displaystyle \arctan{\frac{x}{3}}$$. It should give you a hint what the proper anti-derivative is.)

#### Petrus

##### Well-known member
First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).
Ok now I got pretty unsure. it should be $$\displaystyle \frac{1/3}{1/3}$$ that I factour 1/3 from top and botom

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ok now I got pretty unsure. it should be $$\displaystyle \frac{1/3}{1/3}$$ that I factour 1/3 from top and botom
Note that $$\displaystyle \frac{3^2}{3} \ne 1^2$$.
And if you don't believe me, try calculating both. #### Petrus

##### Well-known member
Note that $$\displaystyle \frac{3^2}{3} \ne 1^2$$.
And if you don't believe me, try calculating both. That make sense.. I believe and can see that.. so I should have $$\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}$$ and it will give me same result with $$\displaystyle \arctan{\frac{x}{3}}$$ is that correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
That make sense.. I believe and can see that.. so I should have $$\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}$$ and it will give me same result with $$\displaystyle \arctan{\frac{x}{3}}$$ is that correct?
No... since $$\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}$$ as well.

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You have $$\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}$$

But what you want is something like $$\displaystyle \frac{1}{(\frac{x}{3})^2+1}$$

#### Petrus

##### Well-known member
No... since $$\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}$$ as well.

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You have $$\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}$$

But what you want is something like $$\displaystyle \frac{1}{(\frac{x}{3})^2+1}$$
Do you mean
$$\displaystyle \frac{1}{(\frac{3x}{3})^2+1}$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Do you mean
$$\displaystyle \frac{1}{(\frac{3x}{3})^2+1}$$
Huh? No.

I mean something like:
$$\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$

#### Petrus

##### Well-known member
Huh? No.

I mean something like:
$$\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$
Thanks.
So now we got $$\displaystyle \frac{3}{3^2}\int\frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$ and if we subsitute $$\displaystyle u=\frac{x}{3}$$ we got $$\displaystyle \frac{3}{3^2}\int\frac{1}{u^2+1}$$ so we got $$\displaystyle \frac{3}{3^2}\arctan{u}$$