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### Limit of Area

```
Date: 03/01/98 at 15:44:59
From: Jason
Subject: Geomety POW-Regarding Mathematical Limitsz

Given a square with sides of length 5. Each side of the square is
divided into three congruent parts, and a square is constructed
outwardly on the middle part. The new figure has twenty line segments
determining its boundary. Each of these is further divided into three
congruent parts, and again a square is formed outwardly on each middle
part. The process is continued infinitely. What is the limit
approached by the area of this figure?

I have tried looking up what limit means, but I don't get what the
area is according to the limit. I know the number of squares
```

```
Date: 03/02/98 at 17:30:20
From: Doctor Sam
Subject: Re: Geomety POW-Regarding Mathematical Limits

Jason,

This is a complex problem. Let's hold off on the idea of "limit" and
just compute some values.

The first square has side length 5. The next square has side length 5/
3. The next squares will be one-third smaller, or 5/9, and the next
would be one-third of that, or 5/27.

Do you see the pattern?  The side lengths of the squares are the
numbers

5, 5/3, 5/(3^2), 5/(3^3), ...

These form what are called a geometric sequence. Each number is a
multiple (in this case, 1/3) of the number that preceeds it.

If you know the lengths of the sides, you can find the area of each
size square

5^2, (5/3)^2, (5/(3^2))^2, (5/(3^3))^2, ...,

which can be simplified to 5^2, 5^2/3^2,  5^2/3^4, 5^2/3^6 ...

Okay so far?

Now, you said that you know the number of squares increases by a
then

another 20 = 4*5, then
another 100 = 4*5^2, etc.

You can find the total area as you go along, adding up the areas at
each stage of the construction:

Total Area =        1 square with side length 5
+ 4 squares with side length 5/3
+ 4*5 squares with side length (5/3^2)
+ 4*5*5 squares with side length (5/3^3)
+ ... etc.

Now let's write them all down:

1*5^2 + 4*5^2/3^2 + 4*5*5^2/3^4 + 4*(5^2)*(5^2)/3^6 + ...

Simplify each fraction (but keep them with exponents to show the
pattern):

Total Area = 5^2 + 4*5^2/3^2 +  4*5^3/3^4  + 4*5^4/3^6 + ...

The idea of a limit is that if you keep adding up more and more terms
in this pattern, the numbers you get will all get closer and closer to
some number (called the limit of this series of numbers).

The limit can be computed because this pattern is also a geometric
series, and there is a formula for the sum of such a pattern. The sum
is:

first term
Sum  =   ----------------
1 - common ratio

To see that this is a geometric series, and to discover its common
ratio, notice that after the first term every term is 5/9 = 5/3^2
times the preceeding term.

So the common ratio is 5/9, but you need to skip the first area of 25
because it isn't multiplied by 4. So take the first term as 4*5.

Total Area = 5^2 + 4*5^2/3^2 + 4*5^3/3^4 + 4*5^4/3^6 + ...
= 25  +  20(5/9)  + 20(5/9)^2 + 20(5/9)^3 + ...

so the first term is 20(5/9) and the common ratio is 5/9. The total
area ... the limit as you keep constructing more and more squares ...
is:

20(5/9)          20(5/9)            20(5)
25  + -------  =  25 + -------   =  25 + ------- = 25 + 25 = 50
1 - 5/9           (4/9)               4

I hope that helps.

-Doctor Sam, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Fractals
High School Geometry
High School Sequences, Series
High School Triangles and Other Polygons

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