1
Sections 1.6 Rigorous Definition of Limit
Some recall: We have talked about ”derivative” in Section 1.1/1.2. We have
seen two different interpretations of this concept:
1. physically as instantaneous speed
2. geometrically as the slope of the tangent line.
f(a)
f (x)−f (a)
x−a
f 0 (a) =
df
dx |a
Physics
A function that gives the displacement
of a particle from a fixed origin
The average velocity of the particle
on the time interval [a,x]
The instantaneous speed of the particle
at time a
Geometry
The graph consisting of the point (a, f (a))
The slope of the secant line between
the points (a, f (a)) and (x, f (x))
The slope of the tangent line to
the graph at the point (a,f(a))
(a)
We have also said that f 0 (a) ≈ f (x)−f
for x close to a. (Instantaneous
x−a
speed is the ”limit” of the average speeds of the particle in the time interval
[a,x]). To be more mathematically correct we have been thinking
df
(a) =
dx
f (x) − f (a)
−a }
| x {z
average rate of change
+ E(x, a)
| {z }
Error
and assuming the error term is going to zero. As we have noted in the GPS
problem the difference between these two quantities – the instantaneous and
the average one – is crucial. (The difference between skipping the tip of the
mountain or just crushing into it on an airplane !!). Hence we need to be
more precise .......
Precision will come with a better understanding of the definition of the
limit.
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We said limx→x0 f (x) = L means if x is near x0 then f(x) is near L.
In a way we are playing a 2-person game between you and I, in which I
claim that the limx→x0 f (x) = L.You take the control of the y-axis and I
(the defender) take charge of the x-axis. I ask:
Tell me what near means to you. So you will give me a constant ²
which will determine a distance criteria between f(x) and L and you declare
|f (x) − L| < ² is near enough for you. Then it is my turn. My tactic is to
choose a ”nearness criterion” (δ) so that when x is close to x0 according to
my ”nearness criterion”, the corresponding f(x) is close to L according to
your near. We play the game until one of the two things happens : either
you run out of all possible your near’s you can think of (in which case I was
able to find my ”nearness criterion” every single time and hence prove the
limit is L) or for one of your near’s I got stumped and was not able to find
my ”nearness criterion”, hence you win and the function f does not have the
limit L as x approaches x0 .
In class I have also used the analogy of a soccer game. On the y-axis around
the point L you set a net, and you get to set the ”width”(²) of this net. Right
after you set your net, it is my turn. My job is to define a counter-width
δ, that will help me to set an interval on the x-axis around x0 . For me to
win the game I have to choose the width (δ) of my interval so that when I
put my ball on any point in this interval and shoot it towards the function,
it reflects from f towards y-axis and gets into your net and hence I score.
When I am able to do this for any width of your net, then the limit exists.
If I fail, even for one of your choices, then it doesn’t.
Below I will give the formal definition of limit. While reading the definition
keep on the side of your mind the game above:
Definition: limx→x0 f (x) = L means for every ² > 0 we can find a δ > 0
such that if |x − x0 | < δ then |f (x) − L| < ²
Let’s see two examples below:
3
Example Claim: limx→1 2x = 2. So the game is on. Let say first you
throw at me ² = 1. This means you have opened an ² = 1 unit wide rectangle
around L=2 on the y-axis and challenged me to define my nearness criterion.
I choose δ = 1/3 (don’t pay attention how I choose mine for the time being
just watch how the game is played.). Now let’s see who is the winner of this
round. If I have chosen mine good, for all x values that are 1/3 units away
from 1, (|x − 1| < 1/3 means just that) all the corresponding f (x) = 2x
values distance to 2 should not exceed your ² = 1. Check out the following
graph:
4
3
Out[13]= 2
1
0.5
1.0
1.5
2.0
It looks like I win this round. The part of f (x) = 2x that falls within my
purple window is shorter than the portion in your red window.
We are not done. You have not exhausted all your possible ² choices yet.
Maybe you were too nice to me with your first choice choose a smaller ²,
such as ² = 1/10. I’ll immediately choose my δ = 1/102 check the graph
below again:
2.6
2.4
2.2
Out[9]= 2.0
1.8
1.6
0.8
0.9
1.0
1.1
1.2
1.3
According to the graph I win again. The part within my purple window is
shorter than the part in your red window. If we continue the game like this
soon you will realize that no matter what ² you give me, I can respond with
a δ that works. Once, it is mutually recognized that, for every possible ², I
can counter with a δ we declare that limx→1 2x = 2.
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Let’s see an example where you get to win:


Example Find the limx→1 f (x) where f (x) =
x
x<1
2
x=1

x+2 x>1
The graph of f (x) is given also below.
5
4
3
2
Out[4]=
1
-1.0
0.5
-0.5
1.0
1.5
2.0
2.5
-1
Assume I claim limx→1 f (x) = 2.Now if you start with an ² such as ² = 1.5.
I will easily choose my δ = 1/2 and according to the graph below I will win:
5
4
3
2
Out[6]=
1
-1.0
0.5
-0.5
1.0
1.5
2.0
2.5
-1
BUT if you choose your ² = 0.5 then there is no δ > 0 I can choose. So
limx→1 f (x) does not exits. Observe below couple of my failed attempts:
5
5
4
3
2
Out[8]=
1
-1.0
0.5
-0.5
1.0
1.5
2.0
2.5
-1
Figure 1: δ = 1/2
5
4
3
2
Out[12]=
1
-1.0
0.5
-0.5
1.0
1.5
2.0
2.5
-1
Figure 2: δ = 1/3
5
4
3
2
Out[14]=
1
-1.0
0.5
-0.5
1.0
1.5
2.0
2.5
-1
Figure 3: δ = 1/8
5
4
3
2
Out[18]=
1
-1.0
0.5
-0.5
-1
1.0
1.5
2.0
2.5