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### Term Limits

```Date: 04/05/2017 at 14:40:07
From: Prince
Subject: Limit sum property violation ?

Given the area under the curve

lim delta x -->0 {Sigma(area 1 + area2 + ... infinity)}

Consider breaking it up into separate areas such as
area 1 = f(x1).delta x and so on until infinity:

lim delta x -->0 area 1 + lim delta x -->0 area 2 + ... infinity

Now, limit delta x -->0 area 1 = 0 (let us suppose). Then the summation of
it, and lim delta x area 2, and all the rest, must be equal to 0.

But this is not happening. Why not? There are infinite terms; but this is
just a statement. Can't we prove this logically? Infinity is just an
extension of terms and nothing else.

Why doesn't this hold like the sum of limits of finite functions?

This question has been nagging me for a very long time!

```

```
Date: 04/05/2017 at 21:41:47
From: Doctor Rick
Subject: Re: Limit sum property violation ?

Hi, Prince.

You are assuming that a property proved for a *finite* number of terms
holds for an *infinite* number of terms. Odd things can happens when you
treat an infinite series the way you would treat an ordinary, finite sum.
You'd have to *prove* that an infinite series can be treated this way --
it's not my responsibility to prove that it *can't*!

Furthermore, what you wrote doesn't recognize that the definite integral
is a limit of a sum of a *finite* number of terms as that number
approaches infinity. Instead, you show the limit as delta(x) approaches
zero of an *infinite* series. In other words, you suppose that we first
take the limit of a series as the number of terms increases toward
infinity, and *then* take the limit of the sum as delta(x) approaches
zero.

Let's take a very simple example of a definite integral and show how it
works in some detail.

When the "slices" are of equal width delta(x), that width is related to
the number of slices by

delta(x) = (b - a)/n

I'm going to abbreviate delta(x) as dx because it will show up a lot in
what follows. Solving for n,

n = (b - a)/dx

The integral from a to b of f(x)dx is then

n-1
lim   Sum f(a + k*dx)*dx
dx->0 k=0

Have you seen something like this before? Can you follow it? Notice that
the sum is a finite sum (of n terms); the first term is

f(a + 0*dx)*dx = f(a)*dx

The nth term is

f(a + (n - 1)*dx)*dx = f(a + ((b - a)/dx - 1)*dx)*dx
= f(b - dx)*dx

You see that I am using the value of the function at the left edge of each
slice as the height of the rectangle.

Now I'll take a very simple function:

f(x) = x

Then this integral ...

n-1
lim   Sum (a + k*dx)*dx
dx->0 k=0

... is equal to

n-1
lim   Sum [a*dx + k*(dx)^2]
dx->0 k=0

What is the sum here? Using the limit sum property you started from, we
can write it as

n-1        n-1
Sum a*dx + Sum k*(dx)^2
k=0        k=0

Since a and dx are constants independent of k, the variable in the
summations, we can use the constant-multiple property of limits to write
this as

n-1            n-1
a*dx*Sum 1 + (dx)^2*Sum k
k=0            k=0

The sum of n terms each equal to 1 is just n.

The sum of k is 0 + 1 + 2 + ... + (n - 1), which is an arithmetic series
the value of which is the average of the first and last terms multiplied
by the number of terms:

(0 + (n - 1))/2 * n = n(n - 1)/2

Putting these results back into the limit, we now have

lim   [a*n*dx + n(n - 1)/2 * (dx)^2]
dx->0

This, again, can be written as the sum of two limits:

lim   a*n*dx + lim   n(n - 1)*(dx)^2/2
dx->0          dx->0

=  lim   a*n*dx + lim   [n^2*dx^2/2 - n*(dx)^2/2]
dx->0          dx->0

Now, let's write n in terms of dx. As I showed earlier, it's (b - a)/dx.
And since dx canceled out and no longer appears in the limit, the first
one becomes just the limit of a constant:

lim   a(b - a) = a(b - a)
dx->0

For the second limit, we have:

lim   [(b - a)^2/2 - (b - a)*dx/2]
dx->0

Once again, we apply the limit sum property to break this into two limits.
The first limit does not contain dx, so it's just the constant value
(b - a)^2/2. The second limit is a constant times dx; since dx approaches
0 as dx approaches 0 (obviously!), this limit is zero.

Putting these pieces together, we find that the integral of f(x) = x
from x = a to x = b is

a(b - a) + (b - a)^2/2 = (b - a)(a + (b - a)/2)
= (b - a)(b + a)/2
= b^2/2 - a^2/2

You may have already worked out what the integral should be, following the
rules for integration rather than going back to first principles. If so,
you will realize immediately that we got the correct answer!

But what should interest you most is that we did NOT get an answer of
zero. This limit of a sum, in which each term approaches zero (but the
number of terms increases at the same time), is not zero.

Are you convinced yet that integral calculus really does work?

- Doctor Rick, The Math Forum at NCTM
http://mathforum.org/dr.math/

```

```
Date: 04/06/2017 at 05:37:18
From: Prince
Subject: {}: Comment

Thank you, Dr. Rick.

I understand that applying that technique was wrong -- it runs counter to
the essence of the 'importance of infinity' in math.

Thanks for correcting me.
```
Associated Topics:
High School Calculus

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