Notesale: Turn your study into money

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CHAPTER 3

Limits and Continuous Functions
1
...
(See § 2
...

1
...
Definition of limit (1st attempt)
...
” It means that if you choose values of x which are close but
not equal to a, then f (x) will be close to the value L; moreover, f (x) gets closer and closer to L as x gets
closer and closer to a
...
)
1
...
Example
...

1
...
Example: substituting numbers to guess a limit
...

We first try to substitute x = 2, but this leads to
22 − 2 · 2
0
f (2) = 22 − 4 = 0
which does not exist
...
Table 1 suggests that
f (x) approaches 0
...

3
...
500000
2
...
010000
2
...
600000
0
...
512195
0
...
500125

1
...
500000
0
...
010000
0
...
009990
1
...
009899
1
...
000000

Table 1
...
” (Values of f (x) and g(x) rounded to
six decimals
...
4
...
The previous example
shows that our first definition of “limit” is not very precise, because it says “x close to a,” but how close is
close enough? Suppose we had taken the function
g(x) =

101 000x
100 000x + 1

and we had asked for the limit lim x→0 g(x)
...
000
...

2
...
” In
the end we don’t really know what they mean, although they are suggestive
...
e
...
Here is the definition
...
Go on to the next sections
...

2
...
Definition of limx→a f (x) = L
...

(2) for every ε > 0 one can find a δ > 0 such that for all x in the domain of f one has
(7)

|x − a| < δ implies |f (x) − L| < ε
...
The inequality x| − y| < δ says
that “the distance between x and y is less than δ,” or that “x and y are closer than δ
...
To prove that lim x→a f (x) = L you must assume that
someone has given you an unknown ε > 0, and then find a postive δ for which (8) holds
...

2
...
Show that limx→5 2x + 1 = 11
...

So, if 2|x − a| < ε then we have |f (x) − L| < ε, i
...

if |x − a| < 21 ε then |f (x) − L| < ε
...
No matter what ε > 0 we are given our δ will also be positive, and if
|x − 5| < δ then we can guarantee |(2x + 1) − 11| < ε
...

22

y = f (x)
L+ε
L

How close must x be to a for f (x) to end up in this range?

L−ε

a

y = f (x)
L+ε
L
L−ε
For some x in this interval f (x) is not between L −ε andL
+ ε
...
You need a smaller δ
...
Therefore the δ in this picture is small
enough for the given ε
...
3
...
We have f (x) = x2, a = 1,
L = 1, and again the question is, “how small should |x − 1| be to guarantee |x2 − 1| < ε?”
We begin by estimating the difference |x2 − 1|
|x2 − 1| = |(x − 1)(x + 1)| = |x + 1| · |x − 1|
...
|Here’s a more concrete situation in which ε and δ appear in exactly the same
roles:

Suppose you are given a circle drawn on a piece of
paper, and you want to know its area
...
You have to do
the algebra with inequalities to compute δ when you
know ε, as in the examples in this section
...

In terms of ε and δ this means that you would
expect that no matter how accurately you want to
know the area (i
...
e
...


Area = πR2
...

When you measure the radius R you will make
an error, simply because you can never measure anything with infinite precision
...
Then the size of the error you made is
error in radius measurement = |x − R|
...
The error in your computed value
of the area is
error in area = |f (x) − f (R)| = |f (x) − A|
...
g
...

Here is a trick that allows you to replace the factor |x + 1| with a constant
...
If we do that, then we will always have
|x − 1| < δ ≤ 1, i
...
|x − 1| < 1,
and x will always be beween 0 and 2
...

If we now want to be sure that |x2 − 1| < ε, then this calculation shows that we should require 3|x − 1| < ε,
i
...
|x − 1| < 13ε
...
We must also live up to our promise never to choose δ > 1, so
if we are handed an ε for which 1 ε > 1, then we choose δ = 1 instead of δ = 1 ε
...

3
We have shown that if you choose δ this way, then |x − 1| < δ implies |x2 − 1| < ε, no matter what ε > 0 is
...
We could
therefore say that in this problem we will choose δ to be
δ = min 1, 13ε
...
4
...
Solution: We apply the definition with a = 4, L = 1/4 and
f (x) = 1/x
...

We begin by estimating |f (x) − 1 4| in terms of |x − 4|:
1 1
4−x
|x − 4|
1
|f (x) − 1/4| =
−
=
=
=

|x − 4|
...
To achieve that we again agree not to take δ > 1
...
How large can 1/|4x| be
in this situation? Answer: the quantity 1/|4x| increases as you decrease x, so if 3 < x < 5 then it will never
1
...


1

To guarantee that |f (x) − 4| < ε we could threfore require
1

12 |x

− 4| < ε,

i
...
|x − 4| < 12ε
...
Of course we have
to honor our agreement never to choose δ > 1, so our choice of δ is
δ = the smaller of 1 and 12ε = min 1, 12ε
...
Exercises
43
...

x→2
√
44
...

x→4
√
45
...


37
...

Joe offers to make square sheets of paper for Bruce
...
Bruce asks Joe for a square with
area 4 square foot
...
Bruce doesn’t mind
as long as the area of the square doesn’t differ more than
0
...


x→3

1+x
46
...

2
x→2 4 + x
2− x
47
...

3
x→1 4 − x
x
48
...

x→3 6 − x
√
49
...
However, she needs the error in the area to be less
than 0
...
(She’s paying)
...
Group Problem
...
) Joe is offering to build cubes of
side x
...
For instance, a cube with
2 foot sides has volume+area equal to 23 + 6 × 22 = 32
...
lim 2x − 4 = 6

If you ask Joe to build a cube whose volume plus
total surface area is 32 cubic feet with an error of at
most ε, then what error can he afford to make when he
measures the side of the cube he’s making?

x→1

40
...


41
...
Our definition of a derivative in (7) contains a limit
...
lim x3 = 27
x→3

4
...
” here we describe some possible variations on the concept of limit
...
1
...
When we let “x approach a” we allow x to be both larger or smaller than
a, as long as x gets close to a
...


x→a,x>a

All four notations are in use
...

x%a

x→a−

x→a−0

x→a,x
The precise definition of right limits goes like this:
4
...
Definition of right-limits
...
Then
lim f (x) = L
...

The left-limit, i
...
the one-sided limit in which x approaches a through values less than a is defined in a
similar way
...

4
...
Theorem
...


x→a

In other words, if a function has both left- and right-limits at some x = a, then that function has a limit
at x = a if the left- and right-limits are equal
...
4
...
Instead of letting x approach some finite number, one can let x become “larger
and larger” and ask what happens to f (x)
...

x%∞

(“The limit for x going to infinity is L
...
5
...
The larger you choose x, the smaller its reciprocal 1/x becomes
...

x→∞ x
Here is the precise definition:
4
...
Definition of limit at ∞
...

If there is a number L such that for every ε > 0 one can find an A such that
x > A =⇒ |f (x) − L| < ε
for all x, then we say that the limit of f (x) for x → ∞ is L
...
Instead of δ which specifies how close
x should be to a, we now have a number A which says how large x should be, which is a way of saying “how
close x should be to infinity
...
7
...
To prove that limx→∞ 1/x = 0 we apply the definition to
f (x) = 1/x, L = 0
...

How do we choose A? A is not allowed to depend on x, but it may depend on ε
...

ε
This tells us how to choose A
...
Hence we have proved that lim x→∞ 1/x = 0
...
Properties of the Limit
The precise definition of the limit is not easy to use, and fortunately we won’t use it very often in this
class
...
”
The following properties also apply to the variations on the limit from 4
...
e
...

Limits of constants and of x
...


(P2)

x→a

Limits of sums, products and quotients
...

x→a

x→a

Then
lim F1(x) + F2(x) = L1 + L2,

(P3)

x→a

lim F1(x) − F2(x) = L1 − L2,

(P4)

x→a

lim F1(x) · F2(x) = L1 · L2

(P5)
Finally, if limx→a F2(x)
(P )
6

x→a

0,
lim
x→a

F1(x)
F2(x)

=

L1


...
One can prove these laws using the
definition of limit in §2 but we will not do this here
...

27

There are two more properties of limits which we will add to this list later on
...

6
...
1
...
One has
lim x2 = lim x · x
x→2

x→2

= lim x · lim x
x→2

by (P5)

x→2

= 2 · 2 = 4
...
To apply (P 6) we must check that the denominator (“L2”) is not zero
...

6
...
Try the examples 1
...
4 using the limit properties
...

x→2

x→2

to complete the computation we would like to apply the last property (P6) about quotients, but this would
give us
0
lim f (x) =
...

We have to do something else
...
For such functions there is an algebra trick which always allows you to compute the limit even
if you first get 00
...
In our case we have
x2 − 2x = (x − 2) · x,
so that
lim f (x) = lim

x2 − 4 = (x − 2) · (x + 2)
(x − 2) · x

x


...
) to compute
2
1
=
...
3
...
Of course, you would think that lim x→2 x = 2 and you can
indeed prove this using δ & ε (See problem 44
...
However, if you assume that the limit exists then the limit properties allow us to find this limit
...

x→2

Then property (P5) implies that

√ √
x = lim x · x = lim x = 2
...
We can reject the latter because whatever x
do√
es, its squareroot is always a positive number, and hence it can never “get close to” a negative number like
— 2
...


x gets close to

√

2
...
4
...
Find
√
√
x− 2
lim
x→2
x−2
assuming the result from the previous example
...
e
...
We use the same algebra trick as before, namely we factor numerator and denominator:
√
√
√
√
x− 2
x− 2
1
= √
√ √
√ = √
√
...

x→2
x→2
x−2
4
x+ 2
2 2
6
...
Limit as x → ∞ of rational functions
...


1

=0
x
We even proved this in example 4
...
Using this you can find the limit at ∞ for any rational function R(x) as
in (11)
...
”
x→∞

To find limx→∞ R(x) divide numerator and denominator by xm (the highest power of x occurring in the
denominator)
...


Remember the trick and divide top and bottom by x2, and you get
3x2 + 3
3 + 3/x2
= lim
2
x→∞ 5x + 7x − 39
x→∞ 5 + 7/x − 39/x
limx→∞ 3 + 3/x2
=
limx→∞ 5 + 7/x − 39/x2
3
=
5
Here we have used the limit properties (P∗) to break the limit down into little pieces like limx→∞ 39/x2
which we can compute as follows
lim

2

lim 39/x2 = lim 39 ·
x→∞

x→∞

1
x

2

=

lim 39 ·
x→∞

lim 1
x→∞ x

2

= 39 · 02 = 0
...
6
...
Compute
x
lim

...
This leads to
2
0
x
1/x2
= lim
lim 3
= limx→∞ 1/x 3 = = 0
...
6
...
When limits fail to exist
In the last couple of examples we worried about the possibility that a limit lim x→a g(x) actually might
not exist
...
First let’s agree on what we will call a “failed limit
...
1
...
If there is no number L such that lim x→a f (x) = L, then we say that the limit
limx→a f (x) does not exist
...
2
...
The “sign function1 ” is defined by
sign(x) =

−1
0

for x
<0
for x = 0

1

for x > 0

Note that “the sign of zero” is defined to be zero
...
e
...
But
for arbitrarily small negative values of x one has sign(x) = −1, so one would conclude that L = −1
...
e
...

lim sign(x) does not exist
...
If you think about this formula for a moment you’ll see that sign(x) = x/|x| for all x /= 0
...


−1
Figure 1
...


In this example the one-sided limits do exist, namely,
lim sign(x) = 1 and lim sign(x) = −1
...

7
...
The example of the backward sine
...
e
...

x

When x = 0 the function f (x) = sin(π/x) is not defined, because its definition involves division by x
...
Then,
taking the sine, we see that sin(π/x) oscillates between +1 and —1 infinitely often as x → 0
...


Figure 2
...


Here again, the limit lim x→0 f (x) does not exist
...
So the limit fails to exist in a stronger way than in the example
of the sign-function
...
In the present
example we see that even the one-sided limit does not excist
lim sin
31

π


...
4
...
The expression 1/0 is not defined, but what about
1
lim ?
x→0 x
This limit also does not exist
...

“Common wisdom” is not always a reliable tool in mathematical proofs, so here is a better argument
...
Namely, suppose that
there were an number L such that
1
lim = L
...

x→0 x
x→0
x→0 x
On the other hand 1x · x = 1 so the above limit should be 1! A number can’t be both 0 and 1 at the same
time, so we have a contradiction
...

7
...
Using limit properties to show a limit does not exist
...
Although it is perhaps not so obvious at first
sight, they also allow you to prove that certain limits do not exist
...
Here is another
...
You can turn this around and say that if lim x→a g(x) + h(x) does not exist then either lim x→a g(x) or
limx→a h(x) does not exist (or both limits fail to exist)
...

7
...
Limits at ∞ which don’t exist
...


x→∞

One can prove this from the limit definition (and see exercise 72)
...

L = lim
x→∞
x+2
Once again we divide numerator and denominator by the highest power in the denominator (i
...
x)
x+2− 1
x
L = lim
x→∞ 1 + 2/x
Here the denominator has a limit (’tis 1), but the numerator does not, for if limx→∞ x + 2 − 1 xexisted then,
since limx→∞(2 − 1/x) = 2 exists,
lim x = lim

x→∞

x→∞

x+2 —

would also have to exist, and limx→∞ x doesn’t exist
...
In this situation the limit L itself can never exist
...

8
...
”)
The difference between these two kinds of variables is this:
• if you replace a dummy variable in some formula consistently by some other variable then the value
of the formula does not change
...

• the value of the formula may depend on the value of the free variable
...
The limit is easy to compute:
lim 2x + 1 = 2a + 1
...
This computation says that if some number gets close to a then two times
that number plus one gets close to 2a + 1
...
But the result of our
computation shouldn’t depend on the name we choose, i
...
it doesn’t matter if we call it x or u
...
Some prefer to call x a
bound variable, meaning that in
lim 2x + 1
x→a

the x in the expression 2x + 1 is bound to the x written underneath the limit – you can’t change one without
changing the other
...
For instance, let’s try
setting x = 3 in our limit, i
...
what is
lim 2 · 3 + 1 ?
3→a

Of course 2 · 3 + 1 = 7, but what does 7 do when 3 gets closer and closer to the number a? That’s a silly
question, because 3 is a constant and it doesn’t “get closer” to some other number like a! If you ever see 3
get closer to another number then it’s time to take a vacation
...
For instance, if we set a = 3 (but leave x alone) then we get
lim 2x + 1
x→3

and there’s nothing strange about that (the limit is 2 ·3 + 1 = 7, no problem
...
In general you get 2a + 1
...
Limits and Inequalities
This section has two theorems which let you compare limits of different functions
...
(P6 ) from§5
...
e
...

The first theorem should not surprise you – all it says is that bigger functions have bigger limits
...
1
...
Let f and g be functions whose limits for x → a exist, and assume that f (x) ≤ g(x)
holds for all x
...

x→a

x→a

A useful special case arises when you set f (x) = 0
...

The statement may seem obvious, but it still needs a proof, starting from the ε-δ definition of limit
...

Here is the second theorem about limits and inequalities
...
2
...
Suppose that
f (x) ≤ g(x) ≤ h(x)
(for all x) and that
lim f (x) = lim h(x)
...

x→a

x→a

x→a

The theorem is useful when you want to know the limit of g, and when you can sandwich it between
two functions f and h whose limits are easier to compute
...
The inequalities f≤ g h≤ combined with the circumstance that f and h have the same
limit are enough to guarantee that the limit of g exists
...
Graphs of |x|, —|x| and x cos xπ for —1
...
2
35

9
...
Example: a Backward Cosine Sandwich
...
For example
1
−|x| ≤ x cos ≤ |x|
x
since the cosine is always between −1 and 1
...

x
Note that the limit limx→0 cos(1/x) does not exist, for the same reason that the “backward sine” did not
have a limit for x → 0 (see example 7
...
Multiplying with x changed that
...
Continuity
10
...
Definition
...

Note that when we say that a function is continuous on some interval it is understood that the domain
of the function includes that interval
...

10
...
Polynomials are continuous
...
To show that you have to prove that
lim P (x) = P (2),
x→2

i
...

lim x2 + 3x = 22 + 3 ·2
...
e
...
(P 6) from §5 (just as good, and easier, even though it still takes a few lines to write it out – do both!)
P (x)
10
...
Rational functions are continuous
...
e
...
Then one has
P (x)
lim R(x) = lim
x→a
x→a Q(x)
limx→a P (x)
=
property (P6)
limx→a Q(x)
P (a)
=
P and Q are continuous
Q(a)
= R(a)
...

10
...
Some discontinuous functions
...

For instance, the sign function g(x) = sign(x) from example ?? is not continuous at x = 0
...

36

10
...
How to make functions discontinuous
...

In other words, we take a continuous function like g(x) = x2, and change its value somewhere, e
...
at x = 3
...


x→3

The reason that the limit is 9 is that our new function f (x) coincides with our old continuous function g(x)
for all x except x = 3
...

10
...
Sandwich in a bow tie
...
Consider
(
x cos 1x
for x /= 0,
f (x) =
0
for x = 0
Then f is continuous at x = 0 by the Sandwich Theorem (see Example ??)
...
Since this limit is zero, f (0) = 0
is the only possible choice of f (0) which makes f continuous at x = 0
...
Substitution in Limits
Given two functions f and g one can consider their composition h(x) = f (g(x))
...


x→a

Suppose that you can find the limits
L = lim g(x) and lim f (u) = M
...

This is in fact a theorem:
11
...
Theorem
...


x→a

u→L

Another way to write this is
lim f g(x) = f lim g(x)
...
2
...
The given function is the composition of two functions, namely
√
√
x3 − 3x2 + 2 = u, with u = x3 − 3x2 + 2,
or, in function notation, we want to find limx→3 h(x) where
√
h(x) = f (g(x)), with g(x) = x3 − 3x2 + 2 and g(x) =
x
...


You get the first limit from the limit properties (P1)
...
The second limit says that taking the square
root is a continuous function, which it is
...
3
...

Normally, you write this whole argument as follows:
√
q
√
lim x3 − 3x 2 + 2 = lim x3 − 3x 2 + 2 = 2,
x→3

√

x →3

where you must point out that f (x) = x is a continuous function to justify the first step
...

12
...


65
...


lim (2x + 5)

66
...

54
...

56
...


lim (2x + 5)

x→−∞

69
...


lim (x + 3)2007

x→−4

lim (x + 3)2007

70
...

If limx→0 f (x) and limx→0 g(x) both do not exist,

x→−∞

2
t —2
58
...
True or
false?

t2 + t — 2
t2 — 1

71
...

If limx→0 f (x) and lim x→0 g(x) both do not exist,
then limx→0(f (x)/g(x)) also does not exist
...
lim
t→−1
t2 — 1
61
...
If lim x→a f (x) exists then f is continuous at x = a
...
What are the coordinates of the points labeled A,
...


x→7−

59
...
Group Problem
...
Show
that this implies that lim x→∞ x does not exist
...
Apply the
limit properties to limx→∞ x · 1
...
Evaluate lim

...


5
62
...
lim x 5 + 1
x→∞ x + 2
(2x + 1)4
64
...
Since OAB has area
we find that
θ < tan θ
for 0 < θ < π
...

θ&0
θ
This is a one-sided limit
...

cos θ <

13
...
An example
...

2

This follows from sin 2 θ + cos2 θ = 1
...

1 + cos θ
θ
We have just shown that cos θ → 1 and

sin θ

θ

→ 1 as θ → 0, so (16) follows
...
Exercises

Find each of the following limits or show that it does not
exist
...

sin 2α
79
...
lim

...
lim tan θ
...
lim

...

83
...


87
...
lim
x→0

1 — cos x


...

tan3 x
sin(x2)
lim

...
x→0 1 — cos x
x→0

90
...

cos x

lim
x→π/2

91
...
lim
x→0

cos x
2

x +9

π
2

) tan x
...


93
...

x→π x — π
sin x
94
...

x→0 x + sin x

2x3 + 3x2 cos x

...
lim

lim 1 — sin θ
θ — π/2

lim

x

x→0

θ→π/2

85
...

x
96
...

95
...

39

(!!)
(!! again)

1

2

tan θ

x→∞

x

40

Here is a picture of A12, B6 and π:

97
...

is continuous? If so, find it; if not, say why
...
Find a constant A so that the function
sin x
for x 0
f (x) =
2x
A
when x = 0
99
...
(Hint:
x
x
substitute something)
...
Group Problem
...

(a) Show that An < π < Bn
...
See also:

(b) Show that
n
2π
π
A = sin
and B = n tan
n
n
2
n
n

http://www-history
...
st-andrews
...
uk/
HistTopics/Pi_through_the_ages
...


41

tan θ

sin θ
1
x

74
...
Find a constant k such that the function
(
3x + 2 for x < 2
f (x) =
x 2 + k for x ≥ 2
...

1

— √2

...
A function f is defined by
75
...
Hint: Compute the one-sided limits
...
Find constants a and c such that the function

for x < —1
for — 1 ≤ x < 1
for x ≥ 1
...
The function f is continuous
...


is continuous for all x
...
Two Limits in Trigonometry
In this section we’ll derive a few limits involving the trigonometric functions
...


and

We will use these limits when we compute the derivatives of Sine, Cosine and Tangent
...
2
...
lim

θ→0

sin θ
= 1
...
The proof requires a few sandwiches and some geometry
...
Since the wedge OAC contains the triangle OAC its area
must bewedge
larger
...


θ&0

π

...


√
lim cos θ = lim 1 − sin2 θ = 1
...




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Title: Limits and Continuous Functions
Description: Limits and Continuous Functions ,Informal definition of limits ,The formal, authoritative, definition of limit ,Propagation of errors – another interpretation of ε and δ , Variations on the limit theme ,Properties of the Limit , Exercises ,Examples of limit computations ,When limits fail to exist .Limits and Inequalities

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