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Chapter I V
Functions, Limits and Continuity
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Limits of functions
Let's begin by discussing somethings about the following function:
xxxf sin)( =
Look what happens as we zoom in on a specific region!
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Let's turn our attention now toward limits.
What happens tox
xsinwhenx goes to zero? Does the limit of
x
xsinwhenx goes to 0
exist? If it does exist what is the limit? Doesf(0) exist?
Let's look at another function:
=x
xg1
sin)(
Can we zoom in and make it look any better?
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We can't zoom in aroundx=0. It never will look nice. What is going on here? What
is this function doing whenx gets close to zero? Does the limit of
x
1sin exist when
x goes to 0? Doesf(0) exist?
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We shall define the notation of the limit of a real valued function at a point.
When we say that a functionfapproachesL atxo , we mean that the valuef(x) will lie
in an arbitrary neighborhood ofL provided that we takex sufficiently close toxo.
L +
L
L -
(xo)
xo- xo+
Definition 4.1:
We say that the numberL is the limitoff(x) asx approachesxo and write
Lxfoxx
=
)(lim provided that for each > 0there exists a small positive number > 0
such that
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Example 4.2:
(a) Prove that bbcx
=lim .
Let > 0. We want to show that > 0 such that
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Theorem 4.4: (Sequential Criterion)
Let RA and let RAf : , Rxo . Then the following are equivalent:
(i) Lxfoxx
=
)(lim
(i) For every sequence (xn) inA that converges toxo such that on xx Nn , the
sequencef(xn) converges toL. (i.e., ( ) ( ) Lxfxx non )( ).
Proof: (i) (ii): Assume that Lxfoxx
=
)(lim , and suppose that (xn) is a sequence inA
that converges toxo such that on xx Nn (we want to show that the sequence
f(xn) converges toL). Let > 0.
Since Lxfoxx
=
)(lim , then > 0 such that 0 a pointxA with
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Let 1 = 1, then a pointx1A with 10 1
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(a) If Lxfoxx
=
)(lim and Mxgoxx
=
)(lim , then
MLxgfoxx
+=+
))((lim MLxgfoxx
=
))((lim
LMxfgoxx
=
))((lim bLxbfoxx
=
))((lim
(b) If 0)( xg Ax and 0M , thenM
Lx
g
f
oxx=
)(lim .
Proof: We will prove that MLxgfoxx
+=+
))((lim .
Let (xn) be a sequence inA that converges toxo such that on xx Nn . We want to
prove that the sequence ((f+g) (xn )) converges toL+M.
Lxfoxx = )(lim The sequence (f(xn )) converges toL.
Mxgoxx
=
)(lim The sequence (g(xn )) converges toM.
By theorem (3.15) the sequence (f(xn ) +g(xn)) converges toL+M.
Hence the sequence ((f+g) (xn )) converges toL+M MLxgfoxx
+=+
))((lim .
We can prove the rest of the theorem by using theorems (3.14), (3.16) and (3.17).
Theorem 4.7: (Squeeze Theorem)
Let RA , letf,g and hbe functions onA toR, and let Rxo .
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If )()()( xhxgxf Ax , oxx and if )(lim)(lim xhLxfoo xxxx
== , then Lxgoxx
=
)(lim
Proof: Exercise. ( Use problem 8.5 in H.W. 6)
H.W. 9: P.105 : #8, 9(c), 11(b , d), 13
Continuous Functions
Readings:
Elementary Analysis: The Theory of Calculus. (chapter 3)
Introduction to Real analysis (chapter 5)
In everyday speech, a 'continuous' process is one that proceeds without gaps of
interruptions or sudden changes. Roughly speaking, a functiony = f(x) is continuous
if a small change inx produces a small change in the corresponding valuef(x). We
can also say that a continuous function is a function whose graph can be drawn
without lifting the chalk from the blackboard.
Example 4.8:
Which of the two functions is continuous and which is not:
1. f(x) = 1 ifx > 0 andf(x) = -1 ifx < 0. Is this function continuous ?
2. f(x) = 5x - 6. Is this function continuous?
Since a function is continuous if its graph can be drawn without lifting the pencil, we
will look at the graph of each function:
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f(x) = -1 ifx < 0 and 1 ifx > 0: Since we have to lift the pencil to draw this graph,
this function does not appear to be continuous.
f(x) = 5x - 6: Since this graph, being a straight line, does not require us to lift the
pencil, we would call this function continuous.
Now we give the formal definition of continuity:
Definition 4.9:
Let RA and let RAf : , Axo . We say thatfis continuous atxo
provided that for each > 0there exists a small positive number > 0 such that
Ax and
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f(xo) -
(xo)
xo- xo+
Notes: (1) depends on and onxo.
(2) A comparison of definition (4.1) and (4.9) shows thatfis continuous atxo if and
only if )(lim)( xfxfoxx
o =
Theorem 4.10: (Sequential Criterion for Continuity)
A functionf:AR is continuous at the pointxoA if and only if for every
sequence (xn) inA that converges toxo, the sequence (f(xn)) converges tof(xo).
Proof: Exercise. (See the proof of theorem 4.4) .
Discontinuity Criterion:
A functionf:AR is discontinuous at the pointxoA if and only if there
exists a sequence (xn) inA such that (xn) converges toxo, but the sequence (f(xn)) does
not converge tof(xo).
Definition 4.11:
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LetA R and letf:AR. IfBA, we say thatfis continuous on the set B iff
is continuous at every point ofB.
Example 4.12:
Letf(x) = 2x
2
+ 1. Prove thatfis continuous onR by:
(a) Using definition (4.8).
(b) Using the sequential criterion.
Solution:
(a) Let xoR, and let > 0. We want to prove that > 0 such that if
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Take })12(2
,1min{+
=o
x
.
Then if 0 such that if
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Example 4.14 :
Let
>=
01
0)(
2
x
xxxf . Prove thatfis discontinuous at 0.
Solution: We will use the discontinuity criterion.
Let (xn) =
n
1
, then (xn) converges to 0.
However, the sequence (f(xn)) =
2
1
ndoes not converge tof(0) = -1.
Therefore,fis discontinuous at 0.
Theorem 4.15:
LetA R and let f:AR be continuous at the pointxoA, then f is
continuous atxo.
Proof: Let (xn) be a sequence inA such that onn xx =lim . We want to prove that
)()(lim onn
xfxf = .
fis continuous atxo )()(lim onn xfxf=
.
Let > 0, then then Nsuch that .
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But .
Therefore, )()(lim onn
xfxf = , which means that f is continuous atxo.
Theorem 4.16:
LetA R andf,g:AR be continuous at the pointxoA and let kR. Then
(i) kfis continuous at xo.
(ii) f gis continuous at xo.
(iii) fgis continuous at xo.
(iv)g
fis continuous at xo if 0)( oxg .
Proof:: Let (xn) be a sequence inA such that onnxx =
lim .
fis continuous atxo )()(lim onn xfxf = .
gis continuous atxo )()(lim onn xgxg = .
Apply theorems (3.14) ~ (3.18) to get the results.
Example 4.17:
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Polynomial functions are continuous onR
P(x) = anxn + an-1x
n-1 +----------------------+ a1x + ao.
Theorem 4.18:
LetA,BR, and letf: AR andg:BR be functions such that f(A) B. Iff
is continuous atxoA andgis continuous atf(xo), then the composite function fg
is continuous atxo.
Proof: Let (xn) be a sequence inA such that onnxx =
lim .
fis continuous atxo )()(lim onn xfxf = .
gis continuous at f(xo) )()(lim))(())((lim onnonn xfgxfgxfgxfg ==
.
Hence, fg is continuous atxo.
H. W. 10: P.106: # 8 , 9, 10. P.107: # 3 , 7 , 10 , 11
P. 108 :# 17.8(a,c) , 17.9(a,c) , 17.10(a,b,c)
Properties of Continuous Functions.
Definition 4.19:
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If a constantMsuch that Mxf )( for allx in the domain off, we say thatfis
bounded above and callMan upper bound.
If a constant m such that mxf )( for allx in the domain off, we say thatfis
bounded below and callMan lower bound.
We say thatfis boundedif it is bounded above and below. This is equivalent to
saying that a constantMsuch that Mxf )( for allx in the domain off.
Definition 4.20:
Ifxo is a point in the domain offsuch that )()( oxfxf for all otherx in the
domain, thenfis said to havean absolute maximum atx =xo.
Ifxo is a point in the domain offsuch that )()( oxfxf for all otherx in the
domain, thenfis said to havean absolute minimum atx =xo.
Theorem 4.21: (Boundedness Theorem)
LetI= [a, b] be a closed bounded interval and letf:IR be continuous onI.
Thenfis bounded onI.
Proof: Assume thatfis not bounded onI. Then for each Nn [ ]baxn , such that
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f(xn) > n. So we have a sequence (xn) [a , b] which implies that (xn) is bounded.
Therefore, by Bolzano-Weierstrass theorem (3.41), (xn) has a subsequence )( knx that
converges toxo [a , b], i.e., onkxx
k=
lim .
Sincefis continuous atxo, we must have )()(lim onkxfxf
k=
.
Butf(xn) > n for each Nn += )(lim knk xf which is a contradiction.
Therefore,fis bounded onI.
Remark:
A continuous function on a setAdoes not necessarily have an absolute
maximum or an absolute minimum on the set. For example,x
xf1
)( = does not have
an absolute maximum or an absolute minimum on the setA = (0 , ) orA = (0 , 1).
While it has an absolute maximum and an absolute minimum on the set [1 , 2].
Theorem 4.22: (Maximum Minimum Theorem)
LetI= [a, b] be a closed bounded interval and letf:IR be continuous onI.
Thenfhas an absolute maximum and an absolute minimum onI. i.e. Iyx oo ,
Such that )()()( oo yfxfxf Ix .
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Proof: We will prove thatfhas an absolute maximum.
By the boundedness theorem (4.21) ,fis bounded from above. Let S={f(x):x[a,b]},
then Sis bounded above and by the completeness property c =sup Sexists.
We are going to prove that cS, i.e., c =f(yo) for someyo[a,b]. This will complete
the proof since from the definition ofSand c we have )()( oyfcxf = Ix .
To prove that cS: Assume that cS, then using problem 10.7 Page 71, there exists
a sequence {f(xn)}in Ssuch that {f(xn)} converges to c. But the Bolzano-Weierstrass
theorem tells us that there exists a subsequence )(kn
x that converges to someyo [a ,
b], i.e., onk yx k =lim .
Sincefis continuous atyo, we must have )()(lim onkyfxf
k=
.
But, { )(kn
xf } is a subsequence of {f(xn)} cxf knk=
)(lim by theorem (3.38).
Therefore c = f(yo) and hence cS.
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Similarly, we can prove thatfhas an absolute minimum.
The Intermediate Value Theorem: Proof and Applications
Example 4.23: (Application of the Intermediate Value Theorem)
The population (in thousands) of a colony of bacteria tminutes after the
application of a toxin is given by the function
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(a) The colony dies out when -8t+ 66 = 0 which means t 25.88
66=
Therefore, the colony dies out in 8 minutes and 15 seconds.
(b) Sincep (2) = 5000,p(7) = 10000 andp is continuous on (2 , 7), then the
population was 9000 at some time after 2 minutes and before 7 minutes. Therefore,
there is some time t= kbetween 2 and 7 such that p (k) = 9000.
Theorem 4.23: (The Intermediate Value Theorem)
LetIbe an interval and letf:IR be continuous onI. Ifa ,b I and if Rk
Satisfiesf(a) < k
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(i) Supposef(c ) < kand let = k - f(c ) .
fis continuous at c, then > 0 such that for allx[a , b]
)()()( cfkcfxfcx
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So by the intermediate value theorem there is a numberc(0 , 1) such that
02 =cce .
Example 4.25:
Letfbe a continuous function mapping [0,1] into [0,1]. Show thatfhas a fixed
point. i.e., there is a pointx*[0,1] such thatf(x*)=x*
Solution: The graph offlies in the unit square across the liney=x.
Letg(x) = f(x) x. Thengis continuous on [0,1].
g(0) =f(0) 0 0 andg(1) =f(1) 1 0.
Hence, there is a pointx*[0,1] such thatg(x*)=0f(x*) x* = 0 f(x*) =x*.
H.W. 11: P. 109: # 11, 17
P.110: # 18.5 , 18.7 , 18.8
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Uniform Continuity:
Definition 4.26:
Let RA and let RAf : . We say thatfis uniformly continuous onA
provided that for each > 0there exists a small positive number > 0 such that
Ayx , and
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locations of (of course, choosing a smaller means that I am also allowed to pick
another, smaller - that will work again uniformly for all -locations).
Uniformly continuous
Remarks:
(i) Iffis uniformly continuous on A, then it is continuous at every point ofA.
(ii) It makes no sense to speak of a function being uniformly continuous at a
point.
(iii) While this definition looks very similar to the original definition of
continuity, it is in fact not the same: a function can be continuous, but not
uniformly continuous.
(iv) In the definition of uniform continuity depends only on ,whereas in the
definition of simply continuity depends on and onxo.
Example 4.27:
Let RRf : be defined byf(x) = 2x, and letxoR. Let > 0,2
= .
If =
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Here depends only on and thereforefis uniformly continuous onR.
Example 4.28:
(a) Let RAg : ),0( =A be defined byg(x) =x
1, and letxoA.
Let > 0. We want to find 0> such that
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Example 4.29:
Is the functionf(x) =x2 uniformly continuous on [-7 , 7]?
Let > 0. yxyxyxyfxf +== 22)()( .
1477 =+++ yxyx yxyfxf 14)()( for ]7,7[, yx .
Thus, if14
= then ]7,7[, yx and 0 x ,yIwith
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Then for each nN(lettingn
1= ) xn ,ynIwith
nyx nn
10 0. We want to prove that
(f(xn)) is a Cauchy sequence, i.e. Nsuch that , .
fis uniformly continuous onA > 0 such that
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Ayx , and
