Math 1210-007
Introduction to Limits
Fall 2010
Zeno's Paradox
The following is a slight variation of Zeno's paradox but it is essentially the same idea. We are
studying this in order to develop a more intuitive understanding of limits.
0
1
2
Consider the number line from 0 to 2 as drawn above. Assume that you start at the number zero and
progress towards the number 2 in a series of steps. You take each step in such a way that you cut the
remaining distance to 2 in half. So when you start at 0 the remaining distance is 2, therefore the length
of your step must be one half that distance which is 1. When you are at 1, the distance remaining to 2
is 1, so your next step has a length of ½. Now your distance remaining to 2 is ½, so your third step will
be of length ¼ and so on. The following table summarizes how far you will step and where you will be
located after each step.
Step Number
Step Size
1
1
Total Distance Traveled (Location)
1
2
1
2
1
3
1
4
1 1
1 
2 4
4
1
8
1 1 1
1  
2 4 8
1
n
2n−1
1
2
Define T n to be the total distance traveled so far. Then we see that
n−1
T n =∑ k=0
k

1
2
which is a geometric series with r =
k

n−1 1
1 1
1
1   n−1 =∑k=0
2 4
2
2
1
and the sum going to n – 1 instead of n.
2
In class we proved that a geometric series can be written more concisely by the following formula.
n
S n=∑k=0 r k =
1−r n1
1−r
Therefore we find that the total distance traveled can be written concisely as
n
n−1
T n =∑k=0

1
k
2
1
1
=
=2−
2
1
2
1−
2

1−
n−1

[The last equality requires a few steps of algebra ]
Question: What point on the number line do we appear to be approaching as we take more and more
steps?
Answer: We are getting closer and closer to 2. So it appears that in some sense the point 2 is the limit
for this sequence of steps. In general, when you take a sequence of steps that appear to be getting
closer and closer to a particular point as you take more and more steps, you call that point the limit of
the sequence. So the limit of this particular sequence appears to be 2.
How exactly do we define a limit?
When you can identify a limit for a sequence, you normally call it L. Thus for the sequence T n we
suspect that L = 2.
The reason that L = 2 for the sequence T n is because we get closer and closer to 2 as we take more
and more steps. That is, if you give me any number a0 , no matter how small, I find find a
number n large enough so that ∣T n−L∣a . To see this we will make a few substitutions and
simplifications.
∣   ∣∣  ∣ 
∣T n−L∣= 2−
1
2
n−1
−2 = −
1
2
n−1
=
1
2
n−1
So we need to show that if we choose n large enough, we can satisfy the following inequality
regardless of how small a is:
n−1

1
2
a
Using the properties of logarithms we can solve this inequality for n [Challenge: try to work this out on
your own] and we end up with the following inequality:
n−
loga
1
log2
What is important to notice here is that I have an explicit formula for finding how big the value of n
needs to be for a given value of a. It must be larger than the value on the right side of the above
inequality. For example, let's say I set a=10−50 which is an extremely small number. I want you to
figure out how many steps it will take to reach a point where you are located less than 10−50 units
away from the limit L = 2. Using the formula we derived we can calculate how big n needs to be.
log a
log 10−50 
−50
n−
1=−
1=−
1≈167.1
log 2
log 2
log  2
Since n must be a whole number and must also be greater than 167.1, then clearly we need n to be at
least 168. So after 168 steps we will be within 10−50 units of the limit L = 2. Notice that we could
have found a sufficiently large n regardless of how small a was set to be. Therefore, L = 2 satisfies the
condition for being a limit of the sequence T n .
Calculating a Limit
The above argument is a proof for why the limit of this sequence T n is L = 2. However, if it
was simply our task to calculate this limit, without actually proving it, we could have done this much
more simply. Let's recall our simplified form of T n :
T n =2−
n−1

1
2
n−1

1
Notice that as n gets really really big, the term
gets really really small. This term becomes
2
so small in fact that for large enough n it is essentially equal to 0. In mathematical notation we write
n−1
n−1
1
1
n

∞
(i.e. “
gets really small as n gets really really big”)
0 as
2
2


Therefore, as n  ∞ we see that
T n =2−
n−1

1
2
 2−0=2
So the limit of this sequence is 2 which is very clear to see once you let the exceedingly small term
vanish to zero and disappear. Using this logic we arrive at the correct result of L = 2. This result of
course matches our geometric intuition about the problem (i.e. that we are approaching 2).