Download Ch2.2 Tutorial b1

Chapter 2.2 – b Tutorial --- Finding Limits Graphically and Numerically
In tutorial we will discuss the formal definition of limit.
Definition of Limit:
Let f be a function defined on an open interval containing c (except possibly at c ) and let L
be a real number. The statement lim f ( x ) = L means that for each ε > 0 there exists a δ > 0
x →c
such that if 0 < x − c < δ , then f ( x ) − L < ε
Let’s spend a few moments on a reading lesson and go over this definition phrase by phrase.
•
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
Recall the definition of a
function. It is a rule that links a number in one set with a unique number in a second set.
•
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
This means that you
can evaluate the function at every point in the open interval.
•
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
Suppose the function is
f (x) =
3
on the open interval ( 0,5 ) . The number 2 is located between 0 and 5, yet
x−2
the function is not defined at 2 because f ( 2 ) =
•
3
= undefined .
0
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
It means that L is a
real number.
•
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
This means that we are
talking about a very small interval containing numbers on either side of c , such that the
distance between any of these numbers and c is less than the very small number δ .
•
Let f be a function defined on an open interval containing c (except possibly at c ) and
let L be a real number. The statement lim f ( x ) = L means that for each ε > 0 there
x →c
exists a δ > 0 such that if 0 < x − c < δ , then f ( x ) − L < ε .
This means that when
we evaluate the function with numbers within δ units of c , the answers will be within ε
units of L .
Example 1:
Given the limit lim ( 3 x − 5 ) = 7 , find δ such that 3 x − 5 − 7 < 0.01 whenever
x→4
0< x−4 <δ.
Solution: What we are looking for is the value of δ such that when we evaluate the function
f ( x ) = 3 x − 5 with numbers within the interval ( 4 − δ ,4 + δ ) , the answers will be
between 7 − 0.01 = 6.99 and 7 + 0.01 = 7.01.
Let’s work this several ways. One way is to start with is with 0 < x − 4 < δ .
0< x−4 <δ
−δ < x − 4 < δ
4−δ < x < 4+δ
Now let manipulate the inequality until the middle portion is 3 x − 12 .
3 ( 4 − δ ) < 3x < 3 ( 4 + δ )
3 ( 4 − δ ) − 12 < 3 x − 12 < 3 ( 4 + δ ) − 12
−3δ < 3 x − 12 < 3δ
3 x − 12 < 3δ = ε
This means that δ =
1
1
0.01
1
ε = ( 0.01) =
=
.
3
3
3
300
Let’s check it by evaluating the function at any number within the interval
1
1 

 4 − 300 ,4 + 300  ≈ [3.996666,4.003333] , say at x = 3.999 and see if the answer is within


0.01 units of the limit 7.
f ( 3.999 ) = 3 ( 3.999 ) − 5 = 6.997 and 7 − 6.997 = 0.003 < 0.01
A graphical way of solving the same problem is to see where the lines y = 7 (the limit),
y = 7 + ε = 7 + 0.01 = 7.01, and y = 7 − ε = 7 − 0.01 = 6.99 intersect the line y = 3 x − 5 .
y = 7.01
y=7
(4, 7)
4−δ
4
y = 6.99
4+δ
f (4 − δ ) = 3 (4 − δ ) − 5 = 7 − ε
12 − 3δ − 5 = 7 − ε
1
3
δ= ε
And
f (4 + δ ) = 3(4 + δ ) − 5 = 7 + ε
12 + 3δ − 5 = 7 + ε
1
3
δ= ε
x
Since ε = 0.01, δ =
1
1
.
(0.01) =
3
300
Let’s graph the function and the limit and show you how, with the zoom button on your
calculator, you can check your answer.
y
8
6
4
2
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
-2
-4
-6
-8
Now suppose that we want to know all the x values such that the difference between the
function f ( x ) = 3 x − 5 and the limit 7 is less than 0.1 (the value for ε ). So let’s set our
vertical window to 7 + 0.0 = 7.1 on top and 7 − 0.1 = 6.9 on the bottom. Then we keep
narrowing the horizon window so that the graph is viewable for all values of x in the window.
y
7.09
7.08
7.07
7.06
7.05
7.04
7.03
7.02
7.01
7
6.99
6.98
6.97
6.96
6.95
6.94
6.93
6.92
6.91
In the above graph I just changed the vertical window from 6.9 to 7.1. I did not yet change the
horizontal window. If you look at the graph, there are a lot of values of x that you cannot project
upward and intersect the graph. Let’s narrow the horizontal window until every value of x can
be projected upward to the graph. Let’s first try narrowing it to [3,5] . You can see it is not
narrow enough.
Using the trace button on your calculator, go to the lowest left point of the graph and you will
see something that looks like ( 3.97, 6.9 ) and do the same thing to the upper right hand corner
and get ( 4.03, 7.1) . This means that if you choose any value of x between 3.97 and 4.03, the
function values will be between 6.9 and 7.1. This means that ε = 0.1 and δ = 0.03 .
Note: This is not an acceptable solution, it will just help you check your answers.
Example 2:
Use the ε − δ definition of a limit to prove lim x 2 = 4 .
Solution:
lim x 2 = 4 can be written as lim x 2 − 4 = 0 .
x →2
x →2
x →2
(
)
We want to show that if 0 < x − 2 < δ , then x 2 − 4 < ε .
We can work backwards to determine our path to the proof:
x2 − 4 < ε
( x − 2 )( x + 2 ) < ε
x−2 <
ε
x+2
Since x is close to 2, then x + 2 is between 3 and 5. Since the reciprocal of 5 is less than the
reciprocal of 4, we can safely say x − 2 <
Therefore, one choice for δ is δ =
even less than
ε
5
<
ε
x+2
.
1
ε
ε . You could just as easily have used because it is
5
6
1
ε.
5
So now we start the proof:
x−2 <δ
( x − 2 )( x + 2 ) <
x + 2 δ < 5δ
1
x 2 − 4 < 5δ = 5 ⋅ ε = ε
5
We could also look at the graphical method and determine where the graph intersects the lines
y = 2±ε .
B
A
2
(
)
A is the point 2 − δ , f ( 2 − δ ) = ( 2 − δ ,4 − ε ) and B is the point
( 2 + δ ,f ( 2 + δ ) ) = ( 2 + δ ,4 + ε ) .
Point A:
(2 − δ )
2
= 4−ε
δ 2 − 4δ + ε = 0
δ=
4 ± 16 − 4ε
= 2± 4−ε
2
Since δ is a very small number, we discard 2 + 4 − ε which is close to 4.
δ = 2− 4−ε = 2− 4−ε .
If ε = 0.01, then δ = 2 − 4 − 0.01 ≈ 0.0025
2
f (2 + δ ) = (2 + δ ) = 4 + ε
4 + 4δ + δ 2 = 4 + ε
δ2 + 4−ε = 0
δ=
−4 ± 16 + 4ε
= −2 ± 4 + ε
2
Since δ is a very small number, we discard −2 − 4 + ε which is close to -4.
δ = −2 + 4 + ε = 4 + ε − 2 = 4 + 0.01 − 2 ≈ 0.0025 .
This means that in the interval ( 2 − 0.0025,2 + 0.0025 ) = (1.9975,2.0025 ) the value of the
function will be within 0.01 units of 4.
Let’s test it with x = 1.9999 .
(1.9999 )
2
− 4 ≈ 0.0004 < 0.01
Let’s check our answer by graphing the function f ( x ) = x 2 and the limit y = 4 and set the
upper vertical window to 4.01 and the lower vertical window to 3.99 and then narrow the
horizontal window until the graph appears for every value of x in the horizontal window.
Graph 1 is a regular graph, no narrowing of windows.
y
8
6
4
2
x
-2
2
4
6
8
Graph 2 has the vertical windows narrowed to [3.99, 4.01] . All this shows is the limit graph,
not the function graph. We will need to narrow the horizontal graph until the function graph
comes into view. Let’s try 1.9 to 2.1 first.
This narrowing at least brought the function graph into view. However, there are a lot of values
of x with no graph above it. Let’s narrow it again to 1.99 to 2.01
Still not enough, so let’s place our trace button on the graph and we get 1.9975 on the left and
2.00249 on the right. Let’s narrow it that much and see what happens.
That does it. The horizontal window is 1.9975 to 2.0025. We want to put it in the form 2 ± δ
and determine δ .
2.0025 = 2 + 0.0025 and 2 − 1.9975 = 0.0025
This tells that if δ = 0.0025 guarantees that ε = 0.01.