94 Chapter 3. Limits
Exercise 3.5
1. For each of the following, find the limit, if it exists.
(a) lim
x→2(x2+3x−4) (b) lim
x→7(x2−5x−8)3 (c) lim
x→0
3x−5 2x+7
2
(d) lim
x→1 x√
x2+1 x+1
(e) lim
x→−225+4x (f) lim
x→−2 x2−4 x2+x−2
(g) lim
x→−2 x2−4
x2−x+2 (h) lim
x→6
x2−6x x2−5x−6
(i) lim
x→5 x2+25
x−5 (j) lim
x→3 x−3 x2+9
(k) lim
x→4
√x−2
x−4 (l) lim
x→2 x−5 2−√
x−1
2. For each of the following f, find lim
h→0
f(x+h)−f(x)
h .
(a) f(x)=4x−13 (b) f(x)= x3 (c) f(x)= 1
x (d) f(x)=√
x
3.6. Continuous Functions 95
• Instead of saying (∗0), we will say
(∗∗) f(x) is close to f(a) ifxis close toa.
Roughly speaking, (∗∗) means that ifxis change fromatoa+ ∆x where∆xis a small number (see the following figure which shows the graph ofy= f(x) where f is a function continuous ata), then the corresponding change iny, denoted by∆y, is small, where
∆y= f(a+ ∆x)− f(a).
Remark∆xis a symbol to denote a small change in x; it doesn’t
mean a product of two numbers∆andx. a
∆x∆y
Figure 3.20
If a function f is undefined ata, it is meaningless to talk about whether f is continuous ata. Condition (∗) means that
(1) lim
x→a f(x) exists;
(2) the limit in (1) equals f(a).
If lim
x→a f(x) does not exist or if lim
x→af(x) exists but does not equal f(a), then f is discontinuous ata.
Example Let f(x)=
−1 ifx<0 0 ifx=0 1 ifx>0
. Determine whether f is continuous at 0 or not.
ExplanationThe function is defined on the left-side and the right-side of 0 as well as at 0. Therefore, we may consider whether f is continuous at 0. In fact, we may consider whether f is continuous at 1 etc., but this is another question.
Solution By the definition of f, we have:
x→0−lim f(x)= lim
x→0−−1=−1 and lim
x→0+f(x)= lim
x→0+1=1.
Since lim
x→0−f(x) , lim
x→0+f(x), it follows that lim
x→0f(x) does not exist. Hence f is not continuous at 0.
-3 -2 -1 1 2 3
-1 -0.5
0.5 1
Figure 3.21
Example Let f(x)=
x2 ifx,0,
1 ifx=0. For each real numbera, determine whether f is continuous or discontin- uous ata.
ExplanationThe domain of f isR. So we may consider continuity of f at any pointa ∈R(that is, whether f is continuous ata).
Solution Consider the two cases wherea=0 ora,0:
96 Chapter 3. Limits
(a=0) Note that f(x)= x2on the left-side and the right-side of 0. Thus we have limx→0f(x) = lim
x→0x2
= 02 Theorem 3.5.1
= 0 , f(0) Therefore, f is not continuous at 0.
(a,0) Note that f(x)= x2on the left-side and the right-side ofa. Thus we have limx→af(x) = lim
x→ax2
= a2 Theorem 3.5.1
= f(a) Therefore, f is continuous ata.
RemarkThe graph of f is shown in Figure 3.22.
-2 -1 1 2
1 2 3 4
Figure 3.22
In the preceding definition, we consider continuity of a function f at a pointa(a real number is considered as a point on the real line). In the next definition, we consider continuity of f on an open interval. Recall that an open interval is a subset ofRthat can be written in one of the following forms:
(α, β) = {x∈R:α <x< β}
(α,∞) = {x∈R:α <x}
(−∞, β) = {x∈R:x< β}
(−∞,∞) = R
whereαandβare real numbers, and for the first type, we needα < β.
Definition LetIbe an open interval and let f be a function defined onI. If fis continuous at everya∈I, then we say that f iscontinuous on I.
Remark
• In the definition, the condition “f is a function defined on I” means that f is a function such that f(x) is defined for allx∈I, that is,I ⊆dom (f).
• SinceI is an open interval, we may consider continuity of f at any pointabelonging toI.
• If there existsa∈Isuch that f is not continuous ata, then f isnot continuous on I.
3.6. Continuous Functions 97
Example In the last example, the domain of f is R. The function is not continuous onR because it is not continuous at 0. For the open interval (0,∞), the function f is continuous at all abelonging to this interval.
Therefore, f is continuous on (0,∞). Similarly, f is continuous on (−∞,0).
Example Let f(x)= 1
x. Show that f is continuous on (0,∞) as well as on (−∞,0).
ExplanationThe domain of f isR\ {0}. Since f is undefined at 0, we can’t consider continuity of f at 0. The domain can be written as the union of two open intervals: (−∞,0) and (0,∞). The question is to show that f is continuous on each of these two intervals, that is, f is continuous at everyain the two intervals. We may also say that f is continuous on (−∞,0)∪(0,∞). However, this terminology will not be used in this course.
We will consider continuity on intervals only, because functions continuous on (closed and bounded) intervals have nice properties (see Intermediate Value Theorem and Extreme Value Theorem below).
Proof For everya∈(0,∞), we have
limx→af(x) = lim
x→a 1 x
= 1
a Theorem 3.5.2
= f(a)
Therefore, f is continuous ata. By definition, f is continuous on (0,∞). Similarly, f is continuous on (−∞,0).
RemarkThe graph of f is shown in Figure 3.23.
-4 -2 2 4
-4 -2 2 4
Figure 3.23
Remark Geometrically, a function f is continuous on an open intervalImeans that the graph of f onI has no
“break”; if we use a pen to draw the graph on paper, we can draw it continuously without raising the pen above the paper.
The following two results give examples of continuous functions. They are just immediate consequences of the corresponding results for limits.
Theorem 3.6.1 Every polynomial function is continuous onR.
ExplanationThe result means that ifpis a polynomial function, then it is continuous onR.
Proof Letpbe a polynomial function. For everya∈R, by Theorem 3.5.1, we have lim
x→ap(x)= p(a), that is, p
is continuous ata. Thus by definition, pis continuous onR.
Theorem 3.6.2 Every rational function is continuous on every open interval contained in its domain.
98 Chapter 3. Limits
ExplanationThe result means that if f is a rational function and ifIis an open interval withI ⊆dom (f), then f is continuous onI. Recall that f can be written in the form f(x)= p(x)
q(x) where p(x) andq(x) are polynomials.
• Ifq(x) is never 0, then dom (f)=R.
• Ifq(x)=0 has solutions, then dom (f) is the union of finitely many open intervals:
dom (f)=R\ {z1,z2, . . . ,zk−1,zk}=(−∞,z1)∪(z1,z2)∪ · · · ∪(zk−1,zk)∪(zk,∞), wherez1, . . . ,zkare the (distinct) solutions arranged in increasing order.
Proof Let f be a rational function, that is, f(x)= p(x)
q(x) wherep(x) andq(x) are polynomials. LetIbe an open interval with I ⊆ dom (f). For everya ∈ I, we havea ∈dom (f) and so by the definition of domain, we have q(a) ,0. Therefore, by Theorem 3.5.2, we have lim
x→af(x)= p(a)
q(a) = f(a), that is, f is continuous ata. Thus by
definition, f is continuous onI.
In the preceding definition, we consider continuity on open intervals. If the domain of a function f is in the form [a,b), we cannot talk about continuity of f atabecause f is not defined on the left-side ofa. Since f is defined on the right-side ofa, we may consider lim
x→a+f(x) and also whether the right-side limit equals f(a).
Definition Letabe a real number and let f be a function defined on the right-side of a as well as ata. If
x→a+lim f(x)= f(a), then we say that f isright-continuousata.
Example Let f(x)=√
x. The domain of fis [0,∞). Using a rule similar to Rule (La20), we get
x→0+lim f(x) = lim
x→0+
√x
= 0
= f(0).
Therefore, f is right-continuous at 0. 1 2 3 4
0.5 1 1.5
2
Figure 3.24
In the above example, f is also continuous at everya>0. Thus, it is “continuous” at everyabelonging to its domain, where “continuous at 0” means right-continuous at 0.
Definition LetIbe an interval in the form [c,d) wherecis a real number anddis∞or a real number greater thanc. Let f be a function defined onI. We say that f iscontinuous on Iif it is continuous at everya∈(c,d) and is right-continuous atc.
Similar to the above treatment, we may also consider continuity of functions f defined on intervals in the form (c,d] or [c,d].
Definition Let a be a real number and let f be a function defined on the left-side ofa as well as at a. If
x→a−lim f(x)= f(a), then we say that f isleft-continuousata.
Definition LetI be an interval in the form (c,d] wheredis a real number andcis−∞or a real number less thand. Let f be a function defined onI. We say that f iscontinuous on Iif it is continuous at everya∈(c,d) and is left-continuous atd.
3.6. Continuous Functions 99
Definition LetIbe an interval in the form [c,d] wherecanddare real numbers andc<d. Let f be a function defined onI. We say that f iscontinuous on I if it is continuous at everya∈(c,d) and is right-continuous atc and left-continuous atd.
Example Let f :R−→Rbe the function given by f(x)=
|x| if −1≤x≤1,
−1 otherwise.
Discuss whether f is continuous on [−1,1].
ExplanationIn defining f, the word “otherwise” means thatif x<−1or x >1; this is because it is given that dom (f)=R. Thus we have f(x)=−1 if x<−1 orx>1.
Solution It is straightforward to check that f is continuous at everya∈(−1,1) and that f is left-continuous at 1 and right-continuous at−1. Thus by definition, f is continuous on [−1,1].
Remark
• Note that f is also defined on the right-side of 1 (for example).
Thus we can also consider the continuity of f at 1. In fact, since
x→1+lim f(x) = −1 and lim
x→1−f(x) = 1, it follows that lim
x→1f(x) does not exist and so f is not continuous at 1.
• Let I be an interval in the form [c,d] or [c,d) or (c,d] and let f be a function defined on anopen interval containing I. Then for everya ∈ I, we may consider whether f is continuous ata.
The above example shows that f may be continuous onI but not continuous at somea∈I.
-1 1
-1 1
Figure 3.25
The following theorem describes an important property of continuous functions on intervals. The proof requires a deep understanding ofreal numbersand is beyond the scope of this course.
Intermediate Value TheoremLet f be a function that is defined and continuous on an intervalI. Then for every pair of elementsaandbofI, and for every real numberηbetween f(a)and f(b), there exists a numberξ betweenaandbsuch that f(ξ)=η.
Explanation In the theorem, the condition “f is a function that is defined and continuous on an interval I” means that “f is a function, I is an interval, I ⊆dom (f)and f is continuous on I”.
• Letx,yandzbe real numbers. We say thatz lies between xandyif (1) x≤z≤yfor the case wherex≤y;
(2) y≤z≤ xfor the case wherey≤ x.
Note that ifx=y, thenzlies between xandymeans thatz=x=y.
• BecauseIis an interval, ifaandbbelong toIanda< ξ <b, thenξbelongs toI also.
• The result means that if f is a continuous function whose domain is an interval, then its range is either a singleton (in this case, f is a constant function) or an interval.
100 Chapter 3. Limits
The following result is also called theIntermediate Value Theorem.
Corollary 3.6.3 Let f be a function that is defined and continuous on an interval I. Suppose that a andb are elements ofI such that f(a)and f(b)have opposite signs. Then there existsξ betweenaandbsuch that
f(ξ)=0.
Explanation The condition “f(a)and f(b) have opposite signs” means that one of the two values is positive and the other is negative.
Proof The result is a special case of the Intermediate Value Theorem. This is because f(a) and f(b) have
opposite signs implies that 0 lies between f(a) and f(b).
In the Intermediate Value Theorem, the assumption that f is continuous cannot be omitted. The following example is an illustration.
Example Let f : [0,2]−→Rbe defined by f(x)=
−1 if 0≤ x≤1, 1 if 1< x≤2.
Note that f(0)=−1 and f(2)=1 have opposite signs. However, there does not exist anyξ∈[0,2] such that f(ξ)=0.
We can’t apply the Intermediate Value Theorem. This is because f is not continuous on [0,2]. Indeed, it is not continuous at 0 since lim
x→0−f(x) and lim
x→0+f(x) are not equal.
0.5 1 1.5 2
-1 -0.5
0.5 1
Figure 3.26
Corollary 3.6.4 Let f be a function that is defined and continuous on an intervalI. Suppose that f has no zero inI. Then f is either always positive inI or always negative inI.
ExplanationThe condition “f has no zero in I” means that the equation f(x)= 0 has no solution inI, that is, f(x) , 0 for all x ∈ I. The conclusion “f is either always positive in I or always negative in I” means that either one of the following two cases is true:
(1) f(x)>0 for allx∈I;
(2) f(x)<0 for allx∈I.
Proof Suppose f takes both positive and negative values inI, that is, there exist a,b ∈ I such that f(a) < 0 and f(b) > 0. Then by the Intermediate Value Theorem (Corollary 1), f has a zero betweenaandbwhich
contradicts the assumption that f has no zero inI.
The above corollary is also called the Intermediate Value Theorem. The following example illustrates how to apply the theorem to solve inequalities.
Example Find the solution set to the inequalityx3+3x2−4x−12≤0.
Solution Letp:R−→Rbe the function given by
p(x)= x3+3x2−4x−12.
3.6. Continuous Functions 101
Factorizing we get
p(x)=(x−2)(x+2)(x+3).
The zeros of the function p are −3,−2 and 2 (and no more). Since p is continuous on R, it follows from the Intermediate Value Theorem that on each of the following intervals, pis either always positive or always negative:
(−∞,−3), (−3,−2), (−2,2), (2,∞).
To determine the sign ofpon each of these intervals, we can just pick a point there and find the value (sign) of pat that point. Taking the points−4,−2.5, 0 and 3, we find that
p(−4)<0, p(−2.5)>0, p(0)<0, p(3)>0.
Thus we have
• p(x)<0 forx<−3;
• p(x)>0 for−3< x<−2;
• p(x)<0 for−2< x<2;
• p(x)>0 forx>2.
The solution set is{x∈R:x≤ −3 or −2≤ x≤2}=(−∞,−3]∪[−2,2].
RemarkThe above steps can be expressed in a compact form using a table:
x<−3 x=−3 −3< x<−2 x=−2 −2<x<2 x=2 x>2
p(x) −
p(−4)<0
0 +
p(−2.5)>0
0 −
p(0)<0
0 +
p(3)>0
The next result describes an important property of functions continuous on closed and bounded intervals. It has many important consequences (for example, see the proof of the Mean Value Theorem in the appendix).
Extreme Value Theorem Let f be a function that is defined and continuous on a closed and bounded interval [a,b]. Then f attains its maximum and minimum in [a,b], that is, there existx1,x2∈[a,b]such that
f(x1)≤ f(x)≤ f(x2) for allx∈[a,b].
ExplanationThe theorem is a deep result. Its proof is beyond the scope of this course and is thus omitted.
The following two examples illustrate that in the Extreme Value Theorem,
• closed intervals cannot be replaced by open intervals;
• the assumption that f is continuous cannot be omitted.
102 Chapter 3. Limits
Example Let f : (0,1)−→Rbe the function given by f(x)= 1
x
It is straightforward to show that f is continuous on (0,1). However, the function f does not attain its maximum nor minimum in (0,1). This is because the range of f is (1,∞); f(x) can be arbitrarily large and it can be arbitrarily close to and greater than 1 but it can’t be equal to 1.
1 1
2 3 4
Figure 3.27
Example Let f : [0,1]−→Rbe the function given by f(x)=
1 ifx=0,
1
x if 0< x≤1.
The function f does not attain its maximum in [0,1]. This is because the range of f is [1,∞); f(x) can be arbitrarily large.
Note that f is not right-continuous at 0 since lim
x→0+f(x)=∞(limit does not exist).
1 1
2 3 4
Figure 3.28
Exercise 3.6
1. Let f(x)=
x2 ifx<1, 1 if 1≤ x<2,
1
x ifx≥2.
(a) Sketch the graph of f forx∈[0,5].
(b) Find all the point(s) inRat which f is discontinuous.
2. Let f(x)= x2+x−2
1−√ x .
(a) What is the domain of f? (b) Find lim
x→1f(x).
(c) Can we define f(1) to make f continuous at 1? If yes, what is the value?
3. Let f(x)=sin1
x.
(a) What is the domain of f? (b) Find lim
x→0f(x).
(c) Can we define f(0) to make f continuous at 0? If yes, what is the value?
4. Let p(x)= x5−x4−5x3+x2+8x+4. It is given that the equation p(x) =0 has exactly two solutions, namely 2 and−1. Use this information to solve the inequalityp(x)>0.
5. Let p(x)= x5−6x4−3x3+5x2+7.
(a) Show that the equation p(x)=0 has a solution between 1 and 2.
(b) It is given that p(x)=0 has exactly one solution between 1 and 2. Is the solution closer to 1 or 2?
Chapter 4
Differentiation
4.1 Derivatives
Consider the curve shown Figure 4.1. It is clear from intuition that the “slope” changes as we move along the curve. AtP0, the slope is very steep whereas atP, the slope is gentle (in this sentence, slope means a piece of ground going up or down).
P0
P Figure 4.1
In elementary coordinate geometry, readers have learnt the concept “slope of a line”. It is a number which measures how steep is the line. For a non-vertical line, its slope is given by
y2−y1 x2−x1
where (x1,y1) and (x2,y2) are two distinct points on the line and the value is independent of the choice of the two points.
Figure 4.2
For curves, we shouldn’t say “slope of a curve” because at different points of the curve, the slopes are different.
Instead we should say “slope of a curve at a point”. Below is how we define this concept.
104 Chapter 4. Differentiation
First we have a curve𝒞and a pointPon the curve. To define the slope of 𝒞atP, take a pointQon the curve different fromP. The linePQis called asecant line at P. Its slope, denoted bymPQ, can be found using the coordinates ofPandQ. If we letQmove along the curve, the slopemPQchanges.
𝒞
Q
P
Figure 4.3
Suppose that as QapproachesP, the numbermPQ approaches a fixed value. This value, denoted bym𝒞,P or simply mP if the curve is understood, is called the slope of𝒞 at P; and the line with slope mP and passing throughPis called thetangent line to the curve𝒞at P.
RemarkThe numberm𝒞,P(if exist) is the unique real number satisfying
(∗) mPQis arbitrarily close tom𝒞,P ifQbelonging to𝒞is sufficiently close to (but different from)P.
In view of the concept “limit of a function at a point” and the notation lim
x→af(x), we may write limQ→P
along𝒞
mPQ =m𝒞,P
to mean that (∗) holds. Below, we will discuss how to find lim
Q→P
along𝒞
mPQby rewriting it as the limit of a difference quotient.
Formula for Slope Suppose𝒞is given byy = f(x), where f is a function; andP(x0, f(x0)) is a point on𝒞.
For any point Qon𝒞withQ, P, itsx-coordinate can be written asx0+hwhereh ,0 (ifh> 0,Qis on the right ofP; ifh<0,Qis on the left ofP). Thus,Qcan be written as (x0+h,f(x0+h)). The slopemPQof the secant linePQis
mPQ = f(x0+h)− f(x0) (x0+h)−x0
= f(x0+h)− f(x0) h
Note that asQapproachesP, the numberhapproaches 0. From these, we see that the slope of𝒞atP(denoted bymP) is
mP = lim
h→0
f(x0+h)− f(x0)
h (4.1.1)
provided that the limit exists.
Remark The limit in (4.1.1) is a two-sided limit. This is becauseQcan approachPfrom the left or from the right and sohcan approach 0 from the left or from the right.
4.1. Derivatives 105
Example Find the slope of the curve given byy= x2at the pointP(3,9).
Solution Put f(x)=x2. By (4.1.1), the required slope (denoted bymP) is mP = lim
h→0
f(3+h)− f(3) h
= lim
h→0
(3+h)2−32 h
= lim
h→0
(9+6h+h2)−9 h
= lim
h→0
6h+h2 h
= lim
h→0(6+h)
= 6.
-3-2-1 1 2 3 2
4 6 8 10 12 14
Figure 4.4
Definition Letx0be a real number and let f be a function defined on an open interval containingx0. Suppose the limit in (4.1.1) exists. Then we say that f isdifferentiableatx0.
Convention Open intervals will be denoted by (a,b) including the cases wherea =−∞and/orb=∞, unless otherwise stated. Thus (a,b) can be any one of the following:
• (a,b) wherea,b∈Randa<b;
• (−∞,b) wherea=−∞andb∈R;
• (a,∞) wherea∈Randb=∞;
• (−∞,∞) wherea=−∞andb=∞.
Remark
• The condition “f is a function defined on an open interval containing x0” means that there is an open interval (a,b) such that (a,b)⊆dom (f) andx0∈(a,b). Hence, fis defined on the left-side and right-side ofx0as well as atx0. The expression f(x0+h)−f(x0)
h in the limit in (4.1.1), considered as a function ofh, is defined on the left-side and the right-side of 0 but is undefined at 0.
• “f is differentiable at x0” means that the slope of the curve𝒞atPexists, where𝒞is given byy = f(x) andPis the point on𝒞whosex-coordinate isx0.
• There is an alternative way to describe the limit in (4.1.1). Puttingx= x0+h, we havex−x0=h. Note that ashapproaches to 0,xapproaches tox0. Hence we have
h→0lim
f(x0+h)− f(x0)
h = lim
x→x0
f(x)− f(x0) x−x0 .
Theorem 4.1.1 Let x0 be a real number and let f be a function defined on an open interval containing x0. Suppose f is differentiable atx0. Then f is continuous atx0.
106 Chapter 4. Differentiation
Proof Since f is differentiable atx0, by definition, lim
x→x0
f(x)−f(x0) x−x0
exists (a real number). Hence we have
x→xlim0
f(x)− f(x0)
= lim
x→x0
f(x)−f(x0) x−x0
·(x−x0)
= lim
x→x0
f(x)−f(x0) x−x0
· lim
x→x0
(x−x0) Limit Rule (La5)
= lim
x→x0
f(x)−f(x0) x−x0
·(x0−x0) Theorem 3.5.1
= lim
x→x0
f(x)−f(x0) x−x0
·0
= 0.
Therefore, we get
x→xlim0 f(x) = lim
x→x0
f(x)− f(x0)+ f(x0)
= lim
x→x0
f(x)− f(x0) + lim
x→x0
f(x0) Limit Rule (La4)
= 0+ f(x0) From above and Limit Rule (La1)
= f(x0).
That is, f is continuous atx0.
The following example illustrates that converse of Theorem 4.1.1 is not true.
Example Let f(x)=|x|. The domain of f isR.
The function f is continuous at 0. This is because
• lim
x→0−f(x)= lim
x→0−(−x)=0;
• lim
x→0+f(x)= lim
x→0+x=0, and so lim
x→0f(x)=0= f(0). -
2 -1 1 2
1 2
Figure 4.5
However, f is not differentiable at 0. This is because lim
h→0
f(0+h)−f(0)
h does not exist as the left-side and right-side limits are unequal:
h→0+lim
f(0+h)−f(0)
h = lim
h→0+
h−0 h
= lim
h→0+1
= 1
and lim
h→0−
f(0+h)−f(0)
h = lim
h→0−
−h−0 h
= lim
h→0−−1
= −1.
Definition Let f be a function.
(1) Suppose that f is differentiable at every point belonging to an open interval (a,b). Then we say that f is differentiable on(a,b).