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1
Chapter 2: Limits of functions:
In this chapter we will cover:
2.1 The tangent and the velocity problems (motivation)
2.2 The limit of a function (intuitive approach)
2.4 The precise definition of the limit of a function
2.3 Limit laws. Limits of elementary functions and the squeeze theorem.
2.6 Limits at infinity and infinite limits . Horizontal and vertical asymptotes.
2.5 Continuity
2.7 Review .
2.1. The tangent and the velocity problems (motivation):
A. The tangent problem:
Consider the graph of a function )(xf , such as the graph shown below (which is the graph of 14)( 3 xxxf ):
Figure 1: Graph of
14)( 3 xxxf
How can we define and find the tangent line to this graph at a point on the curve (say at the point )1,2(P )?
In order to understand how to approach this problem, consider the slightly simpler function: 2)( xxf graphed
below:
Figure 2: Graph of 2)( xxf
2
How can we define and find the equation of the tangent line to this graph at )1,1(P ?
One definition of the tangent line: The tangent line to the graph of a function )(xf at a point ))(,( afaP on the
graph is the line which touches the graph of )(xf only at the point P (if we look only close “enough” to the point)
and which follows the direction of the curve at the point of contact.
How can we obtain a working method from this definition?
Consider two points on the graph of 2)( xxf : )1,1(P and ))(,( xfxQ with 1x .
The points P and Q determine a secant line (a line which intersects the curve more than once) to the graph of )(xf .
Then we determine the slope of the tangent line as PQ
PQmm
lim , where PQm is the slope of the secant line PQ .
In this case (for 2)( xxf ), fill the following table in order to determine the slope m :
Therefore, it appears that 2m in this case, and therefore the equation of the
tangent line to the graph of 2)( xxf at )1,1(P is :
(1) 12121 xyxy
Example :
Following the same method as described above, calculate the equation of the tangent line to the graph of 2)( xxf
at )0,0(P and at )1,1(P .
B. The velocity problem:
As you watch the needle of the speedometer of your car / vehicle, you see that the speed of the car varies in time..
How is this instantaneous speed found at each instant in time ?
Consider that we determine that the position of your vehicle in time is given by the function 29.4)( tts where s(t)
is measured in Kilometers and t is measured in hours. Let us determine the instantaneous speed of this vehicle at
time hour 1t .
Since we do not have (yet) a definition for instantaneous speed, one way to do this is to reme mber that the average
speed of an object over a time interval 21,tt is given by:
12
12, 21
tt
tstsv Avg
tt
(change in position divided
by the change in time over this time interval) and define the instantaneous velocity (or instantaneous speed ) at 1 as
x PQm
3
2
1.5
1.1
1.01
0
0.5
0.9
0.99
3
1
)1()(limlim)1(
1
,1
1
t
stsvv
t
t
Avgt
where t is an arbitrary time 1t .
Fill the following table in order to determine the instantaneous velocity at 1t for 29.4)( tts :
Note: You may remark that the calculations performed in problems A and B are very similar. This is because we can
interpret the instantaneous velocity )1(v (at 1t ) as the slope of the tangent line to the graph of )(ts at the point
)9.4,1(P .
Homework: from Exercise Set 2.1: problems 3,4,5,6 and 8.
2.2 The limit of a function (intuitive approach):
In the previous section we used the notation PQPQmm
lim and t
Avgt
vv ,1
1lim)1(
and these notions were introduced
intuitively. However, what is the limit of a general function )(lim xfax
, and how do we calculate such a limit?
As suggested by the previous examples, our first intuitive definition of )(lim xfax
is:
Definition 1: )(lim xfax
is the value L (if it exists) which )(xf approaches when ax appoaches , but ax .
This definition offers us a simple but non-rigorous method to calculate limits.
Example 1: Consider 2)( 2 xxxf . Calculate )(lim2
xfx
by building a table of value as below, and infer from
this table what the value of this limit should be:
Then graph )(xf to confirm your claim..
Note that the value of )(lim2
xfx
is in general totally unrelated to )2(f ,
although in many particular cases, these two values coincide.
t Time Interval
tAvgv ,1
2 ]2,1[
1.5 ]5.1,1[
1.1 ]1.1,1[
1.01 ]01.1,1[
1.001 ]001.1,1[
x 2)( 2 xxxf
1
1.5
1.9
1.99
1.999
3
2.5
2.1
2.01
2.001
4
Example 2: Calculate 1
1lim
2
1
x
x
x using a table method as above. Note that in this case
1
1)(
2
x
xxf is undefined at
1. In fact, )(xf can take any value at 1x without changing the value of )(lim1
xfx
.
A slightly more precise definition of )(lim xfax
:
Definition 2: We say that Lxfax
)(lim if (all values of) )(xf are (or can be made) arbitrarily close (or as close as
we desire) to L when x is close enough to a.
What this says is that the entire graph of the function )(xf is situated in an arbitrarily narrow band around L , when
x is in a narrow enough band around a.
Example 3: Using the method shown in Examples 1 and 2, and keeping in mind Definition 2, calculate:
x
x
x
)sin(lim
0
Then graph x
xxf
)sin()( to confirm or clarify your answer.
Example 4: Estimate 2
2
0
39lim
x
x
x
Then graph 2
2 39)(
x
xxf
to confirm or clarify your answer.
5
Example 5: Estimate, if possible:
xx
sinlim
0
Example 6: The function
0for ,1
0for ,0
)(
x
and
x
xf is graphed below:
Figure 3: Graph of the function )(xf given above;
What is )(lim0
xfx
? We see that as 0)( : 0 with 0 xfxx , but as 1)( : 0 with 0 xfxx , and we write:
1)(lim and 0)(lim00
xfxfxx
; hence )(lim0
xfx
does not exist in this case, since there is no single number L such
that )(xf is arbitrarily close to L (or which )(xf approaches) when x is close enough to 0 .
Definition 3 :
a) We write )(lim xfax
=L if we can make the values of )(xf arbitrarily close to L when x is close enough to a, but
ax .
b) We write )(lim xfax
=L if we can make the values of )(xf arbitrarily close to L when x is close enough to a, but
ax .
Theorem 1: Lxfax
)(lim )(lim xfax
= )(lim xfax
=L .
Therefore, if (1) )(lim xfax
)(lim xfax
, then )(lim xfax
does not exist.
Condition (1) is one of the easiest (and frequent) way to show that )(lim xfax
does not exist.
6
Example 7: Do Example 7 on page 92 in the textbook.
Homework: Do problems 1,2,4,6,8,10,11,20,23,25 from Exercise Set 2.2
2.4. The precise definition of limit:
The intuitive definition of limit introduced in 1.2 by definitions 1 and 2 is too vague and inadequate in many cases.
This is because the main terms which are used by this definition: )(xf approaches L (or )(xf is arbitrarily close to
L) when x is close enough to a are too vague and need to be clearly defined and quantified.
For example, how can we be sure that 1)sin(
lim0
x
x
x and not 9999999.0
)sin(lim
0
x
x
x ? and so on for
the other limits.
In order to set limits on a rigorous and precise basis, consider the simple function 12)( xxf , for which we want
to determine )(lim3
xfx
. Intuitively (from a graph of )(xf or from a table) we guess that 5)(lim3
xfx
, but how can
we be sure?
Definition 1: (the precise definition of the limit):
We say that Lxfax
)(lim if for any 0 there exist a 0 such that Lxf )( when ax0 .
Note: We see that this definition expresses rigorously the idea of Definition 2 in 1.2 by quantifying the concept of
“arbitrarily close” with and of “close enough” with . We sometimes even say - close ( as we think of as a
very small number, or a challenge required for the function )(xf ).
For the function 12)( xxf , the illustration of Definition 1 is given in Figure 1 below.
Think of as a variable challenge for )(xf , which determines a
changing corresponding .
Figure 1: Illustration of 512lim3
xx
7
In order to rigorously prove that 512lim3
xx
according to definition 1, we need to express 512x in
terms of 3x . But we see that 2
33262512
xxxx , and therefore
2
in this case.
For confirmation, verify that
5122
3 xx , which proves that 512lim3
xx
in terms of the
definition 1.
So, in general, in order to show rigorously that Lxfax
)(lim we need to find for a given such that
ax0 Lxf )( .
Example 2: Show rigorously that 754lim3
xx
, that is find such that 7x 754x
for any challenge .
Draw a graph of 54)( xxf and confirm your finding.
The illustration of the general case Lxfax
)(lim is shown in Figure 2 below:
Figure 2: Illustration of Lxfax
)(lim
8
Example 3: Show rigorously that 9lim 2
3
x
x
Example 4: Use a graphical method (or otherwise, as above) to show that 265lim 3
1
xx
x
We often need to calculate rigorously and find side limits only.
Definition 2:
a) We say that Lxfax
)(lim if for any 0 there exist a 0 such that Lxf )( when axa (that is
when x is close enough to a but at the left of a ) ;
b) We say that Lxfax
)(lim if for any 0 there exist a 0 such that Lxf )( when axa (that is
when x is close enough to a but at the right of a ) ;
Example 5:
Show that 0lim0
xx
(note that there is no left side limit in this case, and therefore we need to use the right side
limit only)
Homework: Do problems 1,3,4,5,7,13,14,19,21,22, 23,24,26,27,29,32,35 and 36 from Exercise Set 2.4
9
2.3. Limit laws. Limits of elementary functions and the squeeze theorem.
A. Limits Laws. Limits of elementary functions:
Goals:
Show an efficient yet rigorous method to calculate limits: show that afxfax
)(lim if )(xf is an elementary
function and if )(af exists .
Solve limits of the form:
0
0 ;
Lesson:
Theorem 1 (Limit Laws):
Consider 212211 and : and : DDaRDgRDf such that MxgLxf
axax
)(lim and )(lim . Then the following
limits exist and are equal to the indicated values:
(1)
even) isn (when 0 and any for L)(lim
any for )(lim
)()(lim
0for )(
)(lim
)()(lim
)()(lim
)()(lim
any for )(lim
LNnxf
NnLxf
MLxgxf
MM
L
xg
xf
MLxgxf
MLxgxf
MLxgxf
RkLkxfk
nn
ax
nn
ax
ax
ax
ax
ax
ax
ax
Proof:
These laws can be shown using the definition 1 in 1.4, and the methods of section 1.4. Some of these proofs are
harder than others, and are shown in Appendix F of the textbook.
Let us prove 1)1( for illustration purposes.
We have that
(2) for any 01 there exist a 01 such that 1)( Lxf when 10 ax
And we have to show that
(3) for any 02 there exist a 02 such that 2)( Lkxfk when 20 ax .
10
Let us consider 0k (if k=0 then the result and its proof are trivial since it says that 00lim ax
) . Take in (2) 21
k
and in (3) choose 12 . Then we see that (3) is true when (2) is true.
Q.E.D.
From now on, we consider properties (1) true and we use them in problems. Using these, calculate:
Example 1:
a) 32lim 2
5
xx
x b)
1
2lim
2
2
3
x
x
x c)
1
1lim
2
x
x
x
We see that the method to calculate limits ultimately amounts now to substituting )(in with xfax when )(af
exists.
We state this clearly in general:
Theorem 2 (substitution):
a) If xPxf n)( is a polynomial of degree n then aPxP nnax
lim .
b) If xQ
xPxf
m
n)( is a rational function (ratio of two polynomials) then
aQ
aP
xQ
xP
m
n
m
n
ax
lim if 0aQm .
c) If )( xf is an algebraic function (a combination of , polynomials and rational functions by using
addition, subtraction, multiplication or composition), then afxfax
)(lim if )(af exists (or, in other words,
if a is in the domain of f(x)) .
Proof: The proof of a) and b) are direct consequences of Theorem 1. The proof of part c) is given in Appendix F in
the textbook (listed as Theorem 8).
Example 2:
Using Theorem 2, calculate the limits:
a) 23
12lim
2
2
x
x
x b) 635lim 23
3
xxx
x .
Do also Example 1 (Page 99, of Section 2.3) in the textbook.
11
The next theorem will help us solve cases of study of the form
0
0 .
Theorem 3: If )()( xgxf for all Ix , except for Ia , then )(lim)(lim xgxfaxax
.
This theorem allows us to solve limits the form
0
0 , by first algebraically simplifying the function (usually by
cancelling an pax factor from both top and bottom of the fraction) and then calculating the limit of the simplified
function.
Example 3: Using Theorems 3 and 2 above, as well as Theorem 1 in 1.2 calculate the limits:
a) 1
1lim
21
x
x
x b)
h
h
h
9)3(lim
2
0
c)
x
x
x
416lim
2
0
d) If )(lim calculate 4 if 28
4 if 4)(
4xf
xx
xxxf
x
. Then sketch a graph of this function to confirm your result.
e) Calculate xx 0lim
f) Calculate x
x
x 0lim
g) 32
132lim
2
2
1
xx
xx
x h)
1
1lim
3
4
1
t
t
t i)
xxxx 20
11lim
j) 32
9lim
2
2
3
xx
x
x k)
xx
xxx
x
2
23
0lim l)
23
1
2
2lim
22 xxxxx
m) x
xx
x
33lim
0 n)
24
11lim
0
x
x
x
B. The squeeze theorem:
New types of limits (including limits of trigonometric functions and combinations of these) can be calculated using
an inequality result involving limits called the squeeze theorem.
This theorem is based on:
Theorem 4: If )()( xgxf when x is near a (except possibly at a ) and MxgLxfaxax
)(lim and )(lim , then :
(4) ML .
Proof: See the proof of Theorem 2 in Appendix F.
Based on this theorem, we can easily prove the following:
12
Theorem 5 (the squeeze theorem):
If )()()( xhxgxf when x is near a (except possibly at a ) and if )(lim)(lim Lxhxfaxax
, then : )(lim Lxgax
.
Proof: Based on above, if Mxgax
)(lim , then LML , which shows that ML . Q.E.D.
The squeeze theorem can be illustrated graphically and in a diagram as below
)( )( )( xhxgxf
L
Figure 1: Illustration of the squeeze theorem. Figure 2: Diagram of the squeeze theorem
This theorem is especially useful in calculating limits of trigonometric functions.
First, based on it, we can prove:
Theorem 6 (limits of trigonometric functions):
If )(xf is a trigonometric function (one of the functions )cot(or )tan( ),cos( ),sin( xxxx ) or a combination (using
addition, subtraction, multiplication or composition of these functions or a combination of these functions and the
algebraic functions mention in Theorem 2) then
(5) afxfax
)(lim if )(af exists (or, in other words, if a is in the domain of f(x)) .
Proof: see the proofs on page 123 of the textbook.
Note: This theorem, together with Theorem 2 above, allows us to do substitution for all “elementary functions”
(algebraic, trigonometric or combinations of these) as long as a is in the domain of the function.
Example 4: Calculate the following limits:
a) 2
0sinlim x
x b)
x
x
x
22
0sinlim
However, the squeeze theorem is used directly for a different type of problems, where direct evaluation of the
function is impossible:
Example 5: Using the squeeze theorem 5, show that 01
sinlim 2
0
xx
x (see Example 11 on page 105).
Homework: Problems 1, 2, 4,9,13,15,17,19,22,25,27,28,29,32,33,36,40,41,42,48,50 and 63 from Exercise Set 2.3 .
13
2.6. Limits at infinity and infinite limits:
So far we are able to calculate afxfax
)(lim if a is in the domain of )(xf , or for a case of study of the form
0
0.
In this section we will study new important types of limits involving .
A. Infinite limits:
Definition 1:
a)
)(lim xfax
if the values of )(xf can be made arbitrarily large when x is close enough to a , or in symbols, if for
any 0A (no matter how large !) there exist a 0 such that Axf )( when ax0 .
b)
)(lim xfax
if the values of )(xf can be made arbitrarily large negative when x is close enough to a , or in
symbols, if for any 0A (no matter how large !) there exist a 0 such that Axf )( when ax0 .
Example 1:
a) Graph 2
1)( and
1)(
xxg
xxf and infer infinite limit results from the graph.
b) Using the definition above, show that
1
lim and 1
lim , 1
lim ,1
lim00
2020
xxxx xxxx
c) What is
xx
1lim
0 ?
Note: We generalize the results from the example above and state:
(1) 0
1 and
0
1 . These two results mean that
0)(lim if
)(
1lim xf
xf axax
and
0)(lim if
)(
1lim xf
xf axax (of course, in these results a could be replaced by
aa or , as needed).
Also
0)(lim xf
ax means that as x approaches a, f(x) approaches 0 but only with positive values (f(x) is positive as x
approaches a), and
0)(lim xf
ax means that as x approaches a, f(x) approaches 0 but only with negative values (f(x) is
negative as x approaches a) .
The result (1) is the most important tool to calculate infinite limits. Therefore, in order to evaluate a limit of the form
0
1,
we evaluate the function, and determine the sign of the denominator (best by using a sign table for it !!), then use (1).
14
Example 2:
Calculate the limits:
a) 3
2lim
3 x
x
x and
3
2lim
3 x
x
x , therefore ?
3
2lim
3
x
x
x
b) 23 3
2lim
x
x
x
c) )tan(lim
2
xx
and )tan(lim
2
xx
, therefore ?)tan(lim
2
xx
d) Do problems 29, 31 and 35 from Exercise Set 2.2 in the textbook.
In the previous examples, if )(or )(lim
xfax
, or if )(or )(lim
xfax
or if )(or )(lim
xfax
, then
the line ax is called a Vertical Asymptote (V.A.) for the graph of )(xf (see, for example, the limits in Example 1).
Therefore:
Definition 2: The line ax is called a Vertical Asymptote (V.A.) for the graph of )(xf if at least one of the following
six cases takes place:
)(or )(lim
xfax
, or )(or )(lim
xfax
or )(or )(lim
xfax
.
Practically, to calculate the V.A. for a given function, we calculate first the zeros of the denominator (if the function
is a fraction) and then calculate the limit of the function at these points.
Example 3:
a) Problem 38 from problem set 2.2 in the textbook.
b) Find all the vertical asymptotes of :
i) 95
8)(
xxxf ii)
4310
7)(
x
xxf
iii)
x
xxf
)cos()(
iv) x
xxf
cos)( v)
x
xxf
)sin()( vi)
x
xxf
2sin
)tan()(
B. Limits at infinity:
We turn now to the (last main) case when x .
Definition 3:
a) Lxfx
)(lim if for any 0 there exist a 0B such that Lxf )( when Bx
(or in words if )(xf is arbitrarily close to L when x is large enough). Alternately:
b) Lxfx
)(lim if for any 0 there exist a 0B such that Lxf )( when Bx
(or in words if )(xf is arbitrarily close to L when x is large enough and negative).
15
c) Definition 1 can be combined with Definition 3 (use this as Exercise) to produce:
)(or )(lim and )(or )(lim
xfxfxx
.
Example 4:
a) Calculate, using the definition above: xx
1lim
and
xx
1lim
b) Graph these functions to confirm your result .
c) Calculate 2
1lim
xx .
Note: We generalize the results from the examples above and state:
(2) 01
This means that
)(lim if 0
)(
1lim xf
xf axax (which is relatively easy to prove using the
Definitions 1. a from this section and Definition 1 in 1.4.
From now on, we will use the result (2) as a main tool to calculate limits at .
Example 5: Calculate the following limits by using (2) above:
a) 2
lim x
x
x b)
23lim
2
2
xx
x
x c)
32
1lim
xx
d) xxxx
39lim 2
e) xxxx
2lim 2
The following general results involving will facilitate the direct substitution of in limits:
(3)
)(lim if ,
)(lim if ,
if ,
xga
xga
Ra
a
x
x (4)
0)(lim if ,0
0 if ,
0 if ,
xga
a
a
a
x
(5)
0)(lim if ,
0 if ,0
0 if ,
0 xgn
n
n
x
n
(6)
)(lim if ,
0 if ,
0 if ,
0 if ,
xga
a
a
a
a
x
16
These results can be proved rigorously with a similar approach as described in (1) and (2). Understand them by
thinking of as a quantity larger than any real number, memorize especially the formulas (1) to (6) , and solve the
following problems by substitution:
Example 6:
a) 4
53lim
x
x
x b)
532
lim
2
3
xx
xxx
x c)
1
9lim
3
6
x
xx
x
d) bxxaxxx
22lim e) 1lim 2
xx
f) 2
3lim
3
24
xx
xxx
x
g) x
xearctanlim
h) xx
xx
x ee
ee33
33
lim
i) xx
xx
x ee
ee33
33
0lim
j) 1
)(sinlim
2
2
x
x
x k) )cos(lim 2 xe x
x
l) 32lim
3
xx
x
m) 3
2lim
3
x
x
x n)
35 5lim
x
e x
x o)
65
82lim
2
2
2
xx
xx
x
p) )cot(lim xx
Definition 4:
a) The line by is called a Horizontal Asymptote (H.A.) at for the graph of )(xf if
bxfx
)(lim
b) The line by is called a Horizontal Asymptote (H.A.) at for the graph of )(xf if
bxfx
)(lim
Therefore, a V.A. involve a finite x value and an infinite f(x), while a H.A. involve an infinite x value and a finite
f(x) value there.
In problems, to calculate all H.A. for a given function, calculate )(lim and )(lim xfxfxx
. If one of these limits
produce a finite value, then the function )(xf has a H.A. there, otherwise it doesn’t.
Example 7:
Find the Horizontal Asymptotes at at and for the following functions. Then graph these functions to
visualize your result:
17
a) 1
1)(
2
2
x
xxf b)
2
12)(
2
2
xx
xxxf c)
56)(
2
3
xx
xxxf
Homework:
What problems are left from above, and problems 19, 21,22,38,41,44 and 45 from the Exercise Set 2.6 in the
textbook.
2.5 Continuity of functions:
Functions which have the property that )()(lim afxfax
are called continuous at a . Continuity of functions is a
very important property which some functions verify on an entire interval or on their entire domain of definition.
Functions which are continuous on an interval will satisfy other theorems (such as the intermediate value theorem)
which allows us to find their zeros.
In this section we will learn tools (definitions and theorems) to establish the continuity of functions at a point and on
an entire interval, as well as the intermediate value theorem for continuous functions.
Definition 1: A function RDf : is continuous at a number Da if (1) )()(lim afxfax
.
Note that (1) actually says 3 things about the function )(xf . It says that:
a) )(af exists ( )(xf is defined at a ) ;
b) )(lim xfax
exists and
c) )()(lim afxfax
.
Remembering the meaning of )(lim xfax
, we realize that (1) represents the intuitive fact that )(xf does not have a
jump (or a break – discontinuity) at a . Therefore, when we trace the graph of )(xf close to a , the graph passes
continuously through the point )(, afa .
Definition 2:
a) A function which is not continuous at a (that is, which fails to satisfy one of the conditions a), b) or c)
above) is called discontinuous at a .
b) A function for which the condition b) above is satisfied, but one of a) and c) are not satisfied is said to have a
removable discontinuity at a , as in this case the function )(xf can be re-defined at a to make it
continuous there.
Example 1: See Example 1 on page 119 in the textbook.
Example 2: Determine the continuity of the following functions at the indicated point:
a) 2
2)(
2
x
xxxf at 2a b) 0at
0for 1
0 if )(
2
a
x
xxxf
18
c) 0at
0for 0
0 if )(
a
x
xx
x
xf . Then graph these functions to visualize your claim.
Note that some functions (such as xxf )( ) are not defined to the left of 0a , and therefore only a side
continuity can be defined there. Side continuity is also useful to define continuity on an entire interval.
Definition 2:
a) We say that RDf : is continuous from the right at a number Da if (2) )()(lim afxfax
b) ) We say that RDf : is continuous from the left at a number Da if (3) )()(lim afxfax
Example 3: Show that xxf )( is continuous from the right at 0a . Can this function be continuous from the
left at 0a ?
Definition 3:
a) A function Rbaf ,: is continuous on the entire interval ba, is it is continuous at each number
bac , and it is right continuous at a and left continuous at b ;
b) A function Rbaf ,: is continuous on the entire interval ba, is it is continuous at each number
bac , ;
Example 4:
Show that 211)( xxf is continuous on 1,1 .
Following the model of this exercise (and remembering Theorems 2 and 6 in 1.3), we realize that all elementary
functions (functions which are combinations of , polynomials, rational and trigonometric functions) are
continuous at each number in their domain. We state this clearly in the next theorem (which uses similar results for
the other types of functions mentioned):
Theorem 1: The following types of functions are continuous at every number in their corresponding domains:
Polynomials, Rational Functions, Root Functions (Irrational Functions), Trigonometric Functions, Inverse
Trigonometric Functions, Exponential Functions, Logarithmic Functions and combinations of these functions (using
+ , - , multiplication, division or composition ).
Example 5:
a) Where is 1
)ln()(
2
x
xxf continuous and where is it discontinuous?
Hint: Find the domain of )(xf first, then use the result from Theorem 1.
b) Do problems 15, 17 and 22 from Exercise Set 2.5 in the textbook.
19
Theorem 2 (The Intermediate Function Theorem):
Let Rbaf ,: be a continuous function on the entire interval ba, , and let u be an arbitrary number strictly
between )(af and )(bf , with )()( bfaf . Then there exists (at least) a number bac , such that ucf )( .
Note: What Theorem 2 suggests is that continuous functions achieve (with a particular )(cf ) all values in their
range (if their range is an interval) . The proof of this theorem can be found in more advanced Calculus texts, and it
is based on Cauchy’s completion axiom (see a model of the proof on Wikipedia).
We Use Theorem 2 to find zeros of a continuous function, or to prove that such zeros exist.
Example 6:
Show that there is a root of the equation 02364 23 xxx between 1 and 2. Can we reduce the length of the
interval which contains this root? Can you suggest an algorithm to find the root to a desired precision?
Homework:
What left from above and problems 3,4,5,6,7, 16, 18, 19, 20, 21,4042,45,46,51 and 54.
2.7 Review:
In this chapter, we learnt
- Limits at an intuitive and motivational level (Sections 1.1 and 1.2) ;
- Rigorous methods to calculate limits (Section 1.4)
- Practical methods to calculate limits (Sections 1.3 and 1.6)
- Continuity of functions based on limits (Section 1.5) .
Learn the main results (definitions, theorems and formulas from these sections) (best write a review guide to include
these) , then solve the following problems:
- In the textbook, do the Review problems from pages 165 to 168, except the problems involving the
derivative of a function or rates of change (problems 11,12,13,14 from Concept Check, problems 19,20,21
from True and False, and problems 38 to 50 from Exercises).
- Do the problems below:
1. Calculate xx
coslim5.0
by building a table of values xx cos, around 0.5 (the intuitive approach of guessing
the limit using a table (used in 1.2) ).
2. Prove the following limits rigorously (using the rigorous definition of limit, as in 1.4):
a) 23
11lim
3
x
x b) 0lim 2
0
x
x c) 154lim 2
2
xx
x
20
3. Calculate the following limits (using methods from the sections 1.3 and 1.6):
a) 5
56lim
2
5
x
xx
x b)
43
4lim
2
2
1
xx
xx
x c)
h
h
h
8)2(lim
3
0
d) 1
1lim
3
4
1
x
x
x e)
8
2lim
32
x
x
x f)
xxxx 20
11lim
g) x
xx
x
11lim
0
h) 20 16
4lim
xx
x
x
i)
x
x
x
11
0
3)3(lim
j) 1
1lim
3
2
xx
x
x k)
12
23lim
x
x
x l)
542
264lim
3
23
xx
xx
x
m) 1
9lim
3
6
x
xx
x n)
1
9lim
3
6
x
xx
x o) xxx
x39lim 2