CY1201 Calculus Lecture Set 1.pdf

7/29/2019 CY1201 Calculus Lecture Set 1.pdf

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CY1201 Calculus(for C.N. Yang Scholars)

Ng Keng MengAY 2013/14


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Chapter 1

Basics (sets, functions, proofs)

1.1 Sets

Often in life and science we would consider collections of objects describing how objects ofdifferent nature correspond to each other. Mathematically, a collection of objects is called a setand the individual objects of the collection we will call elements. The standard mathematicalnotation is x A, which means that the element x belongs to the set A.

DEFINITION 1.1. If each element of the set A belongs to the set B, then we saythat A is a subset ofB. We write A B.

If A B and B A then the two sets are equal, i.e. A = B.If A B and A = B then we write A B. In this case B has strictlymore elements than A.

The most important sets which we will come across in this course are the following.

(i) The set of natural numbersN = {1, 2, 3, 4, 5, }.

(ii) The set of integers

Z = { , 3, 2, 1, 0, 1, 2, 3, }.(iii) The set of rational numbers

Q =

p

q| p, q Z, q= 0

.

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(iv) The set of real numbers R. This is usually depicted as a continuous line, or sometimeswritten as (, ).

(v) The open/closed/half-open interval

(a, b), (a, b], [a, b), [a, b].

This is the set of all real numbers between a and b, possibly including a, b, depending onwhether the interval is closed.

You might think that including or leaving out the end-points of an interval is no big deal,i.e. (a, b) and [a, b] should be not much different. This is completely wrong! The openand closed intervals are very different mathematically and many theorems we study incalculus will only hold for open intervals but fail for closed intervals (or vice versa).

Exercises marked with (such as the one below) are usually more challenging and lessstraightforward than the other exercises. Please do not be discouraged if you do not know

how to solve the exercises. You should discuss amongst yourselves, or research the internetfor possible hints. The important thing is to keep asking questions and try different methods!

Trying out different methods which dont work is often more instructive than learning fromthe solution given to you. Remember that the process is more important than the outcome!

Exercise 1.2.

(i) Which of the sets N,Z,Q and R is larger than the others?

(ii) Are there more rational numbers in Q, or are there more real numbers in the interval(0, 0.00000000001)?

1.2 Functions

A function f assigns a relationship between each element of a set A and an element of a set B.A is called the domain of f and B is called the co-domain of f. A function f always assignsto each element x A a unique image f(x) B. The range of f is the set of all elements ofB which is the image of some x A. In this course we usually deal with functions f : A R,where A R. Thus mathematically, we can define

domain f = {x R | f(x) exists}.range f = {y R | we can find some x such that f(x) = y}.

EXAMPLE 1.3. The map f that takes each x (0, ) to a y such that y2 = x is not afunction, because f(x) can be

x or x. However the map g that takes each x to y = x2

is a function.

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Figure 1.1: The domain is (, ) while the range is (0, 1).

EXAMPLE 1.4. If the domain of f is N then we call f a sequence. The values of f canbe written as f(1), f(2), f(3), . We commonly write an = f(n) and so a sequence is oftenwritten instead as

{an}, {an}n=1, {a1, a2, a3, }A sequence is usually defined explicitly, i.e. an explicit formula is give for an. For instance,the sequence {1, 2, 4, 8, 16, 32, } is given by the formula an = 2n. Can you guess the formulafor the sequence {2, 8, 26, 80, 242, }?

Sometimes a sequence can be defined recursively. This involves specifying the first terma1 and then describing how to obtain an+1 if we are given the value for an. For instance,

a1 = 2, an+1 = 3an

Then we can compute the terms of this sequence 2, 6, 18, 54, . Can you write down theexplicit formula for an?

An important concept is to know how to determine the domain and the range of a givenf. To find the domain of a given f, we usually check for which values of x is f(x) defined. Tofind the range of f we can either read off by inspecting the formula for the function f(x), orwe can set y = f(x) and try to write x in terms of y.

EXAMPLE 1.5. f(x) = 1 x2. This expression makes sense for any value of x R, so thedomain is R. Now as x ranges over the domain R, x2 will range over [0, ), and 1 x2 willrange over (, 1]. Thus the range of f is (, 1].

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EXAMPLE 1.6. f(x) =

1 x2. This expression makes sense only when 1 x2 0, so xmust come from the interval [1, 1]. Thus the domain is [1, 1]. As x ranges over the domain[1, 1], we get 0 1 x2 1 and so 0 1 x2 1. Therefore the range is [0, 1].

EXAMPLE 1.7. f(n) =1

n2 8n + 15 . This expression makes sense whenever n28n+15 =

0. However n2 8n + 15 = (n 3)(n 5) which is equal to 0 if and only if n = 3 or 5. Thusthe domain is any n except for n = 3, 5. Mathematically we write this as N {3, 5}. Whatcan we say about the range? It is not easy to describe so we will not do this here.

EXAMPLE 1.8. Find the domain and range of the function f(x) =x 1x2 + 1

.

SOLUTION . First, lets try and find the domain. When can f(x) fail to exist? There aretwo concerns; we need to ensure that the denominator x2 + 1 = 0, and we need to ensurethat the inside of the square root is non-negative. First, the denominator x2 + 1 1 for anyx, so the denominator is never 0. Next, we need to pick x so that

x 1x2 + 1

0. This is truewhenever x 1 0. Thus, the domain is [1, ).

Lets now find the image. Its too complicated to obtain the range by inspecting the

formula for f(x), so let us set y = f(x) =x 1

x2

+ 1

and find all values of y such that we can

find a correspond x. To do this, lets try and find all y such that f(x) = y has a solution.First of all notice that y cannot be negative (why?).

y =

x 1

x2 + 1

y2 =x 1

x2 + 1

(x2 + 1)y2 = x 1(y2)x2 x + (y2 + 1) = 0

We need to find all y such that this last equation has a solution in x. This last equation is a

quadratic equation in variable x, so we know that this is solvable if and only if the discriminant

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is non-negative:

1 4y2(y2 + 1) 0

4y4

4y2

+ 1 0(2y2 + 1)2 + 2 0

(2y2 + 1)2 2|2y2 + 1|

2

2y2 + 1

2

Since 2y2 + 1 0y2

2 1

2

|y|

2 12

y 2 1

2[Since y 0]

Thus the image of f is

0,

2 12

.

1.3 Properties of functions

DEFINITION 1.9. A function f : A R is increasing if x y f(x) f(y).

A function f : A R is decreasing if x y f(x) f(y).

A function f is monotonic if it is either increasing or decreasing.

Exercise 1.10. Can you give an example of a function which is

(i) increasing.

(ii) decreasing.

(iii) not monotonic.

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(iv) both increasing and decreasing at the same time.

Can you see that A is important? Varying A may change the properties of a function f.Can you give an example of a function f : R

R and A1, A2, A3 such that f : A1

R is

increasing, f : A2 R is decreasing, and f : A3 R is not monotonic?

DEFINITION 1.11. A function f : A R is bounded if there is a bound M > 0such that for any x A, we have |f(x)| < M.

EXAMPLE 1.12. sin x and cos x are bounded. We can take M = 1.

EXAMPLE 1.13. x2 is unbounded for A = R. However x2 is bounded for A = [1, 1].

DEFINITION 1.14. A function f is

1. odd if f(x) = f(x) for every x.2. even if f(x) = f(x) for every x.3. periodic if there is a period T R such that f(x) = f(x + T) for every x.

EXAMPLE 1.15. sin x and cos x are periodic with period T = 2. Are they even or odd?

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Figure 1.2: An even functionFigure 1.3: An odd function

Figure 1.4: A periodic functionFigure 1.5: Not periodic

Exercise 1.16. The function

f(x) =

1, if x Q,0, if x Q.

is periodic with period q for any q Q.

EXAMPLE 1.17. x2 is even and x3 is odd. A polynomial f(x) = a0 +a1x+a2x2 + +anxn

is odd if it contains only odd powers ofx, i.e. a0 = a2 = a4 = = 0. Similarly the polynomialis even if it contains only even powers of x, i.e. a1 = a3 = a5 = = 0.

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EXAMPLE 1.18. Given any function g(x), the function f(x) = g(|x|) is always even.

Exercise 1.19. Can you give an example of a function which is neither even nor odd? Botheven and odd at the same time?

We can similarly talk about a sequence {an} being bounded, increasing, decreasing ormonotonic. More on this will be covered in the next chapter.

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Chapter 2

Limit of a sequence

2.1 Sequence Limits

We wish to define formally what it means for a sequence to have limit L. Informally, itmeans that as n approaches , we have the terms an approaching the number L.

Lets consider the sequence an =n

n + 1.

As n is large, the difference |1 an| becomes small. In fact, the difference (or error) canbe made as small as we like, by choosing n sufficiently large. This is because |1 an| =1 n

n + 1=

1

n + 1. In this case, we write lim

nan = 1. In general, the formal definition is:

DEFINITION 2.1. A sequence {an} has the limit L, and we write

limn an = L, or an L,

if for every > 0, there is a corresponding integer N() such that

whenever n > N() = |an L| < .

If the limit of a sequence {an} exists, we say that the sequence converges, otherwise wesay that the sequence diverges.

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Note that in the definition, we can replace |an L| < with L < an < L + . We canalso replace |an L| < with |an L| , and we can replace n > N() with n N().

Figure 2.1: Pick N() depending on

Generally, a smaller requires a larger N(), as the red band will get slimmer. This iswhy we write N() as a function of , since different will generally require a different valuefor N.

EXAMPLE 2.2. Show by using the definition of the limit that the sequence an convergesto 0, where

an =2n+1

3n

SOLUTION . Let > 0 be given. (This means that the value of > 0 is fixed from this pointon.) We work backwards, starting with the final inequality we need, |an 0| < , and then

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working backwards to make n the subject:

|an

0

|<

2n+1

3n

<

2 2n

3n<

2

3

n ln

2

/ ln

2

3

.

The last inequality is switched around because ln23 < 0. Finally, after making n the

subject, we can read off N() from the right hand side. Usually we would like to take N()to be a positive integer, so we take the rounded up value. So our answer is

N() =

ln2

ln

23

.Now to check that our answer for N() works. This means that we should start with n > N().

Then we have n > ln

2

/ ln

2

3

following the steps above in reverse |an0| 4 n > 3

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Thus we can take N(0.25) = 3. In fact, we can take N(0.25) to be 5, 10, 10023 or any numberlarger than 3. Why?

Repeating for the other values of , we can take N(0.1) = 9 and N(0.001) = 999.

EXAMPLE 2.4. Prove by using the definition of the limit that limna = a. By this we mean

the constant sequence a1 = a2 = a3 = = a.

SOLUTION . Let > 0 be given. Our task is to find N() so that for every large n > N(),we have |an a| < . However the quantity |an a| = |a a| = 0 < for any value of n,whether large or small, and so we can take N() to be anything we want. Lets take N() = 1.

EXAMPLE 2.5. Prove by using the definition of the limit that limn e

n = 0.

SOLUTION . Let > 0 be given. Our task is to find N() so that for every large n > N(),we have |an L| < . So now working backwards,

|an L| < |en 0| <

en <

ln en < ln n < ln n > ln

So we take N() = ln .Why is there a minus sign in ln ? Is there something wrong?

Exercise 2.6.Prove by using the definition of the limit that limn

1

n2 = 0.

Sometimes it is not easy to make n the subject and so it is not clear how to obtain N().A common trick is to replace the expression |an L| by a larger expression and then workwith the larger expression instead.

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EXAMPLE 2.7. Prove by using the definition of the limit that limn

cos n

n= 0.

SOLUTION . Let > 0 be given. Again we work backwards, starting with the final inequality

we need, |an 0| < , and then working backwards.|an 0| <

cos nn

0 <

cos n

n

< At this point we are stuck, because the terms involving n cannot be easily separated tomake n the subject (unlike in Examples 2.2, 2.5). In this case we try and estimate the

expressioncos n

n

from above: We know that | cos n| 1 for every n, and so we see that

cos n

n

1

n

=1

nholds for every n. Now we replace the expression

cos n

n

with the bigger

quantity1

n and proceed as before. (Why can we do this replacement?)

1

n< n > 1

So now we can read off N() =

1

.

Now to verify that our choice of N() works: If n > N() 1n

< . However as1

nis

larger thancos n

n

we can conclude that the smaller quantity is also smaller than , that is,

cos n

n

< . This obviously implies that |an 0| < , which is what we need.

What happens if we try and replace cos nn in the steps above with a smaller quantity?Will the value of N() we get be correct?

Lets try another example where we need to use this trick.

EXAMPLE 2.8. Prove by the definition of limits, that limn

n ln nn +

n

= 1.

SOLUTION . Given > 0 we need to find N(). Lets again start with the end expression:

n ln nn + n 1 <

n + ln n

n + n < Again it looks impossible to separate the terms and make n the subject. Lets try and replace

n + ln n

n +

n

by something bigger and simpler. We know that

n + ln n

n +

n

=

n + ln n

n +

n2

n >4

2

So we may take N() =

4

2

. You should now check that this choice of N() works.

Fact 2.9. In the above example we used the fact that

n > ln n for every natural number n.In fact, for any p > 0, we always have ln n < np for every sufficiently large n. We will provethis fact in a later chapter using the LHopitals Rule. For now you may use this fact withoutproof.

2.2 Properties of limits

We will now talk about several important theorems about the limits of sequences. We willstate these theorems without proving them.

DEFINITION 2.10. A sequence {an} is said to be increasing if an an+1 forevery n 1.

It is said to be decreasing if an

an+1 for every n

1.

A monotonic sequence is a sequence which is either increasing or decreasing.

Note that an increasing sequence does not have to be strictly increasing. In other words,we only require that an an+1 instead of an < an+1. It is possible for a sequence to be bothincreasing and decreasing at the same time. Which sequences can be both increasing anddecreasing at the same time?

In order to show that a given sequence is increasing (or decreasing), there are generallytwo methods:

[Method 1] Prove that an an+1 holds for every n. You can either do this by Mathe-matical Induction, or you can show this directly (See Exercise 2.11).

[Method 2] Use the relationship with functions. Let f(x) be the associated function,i.e. f(n) = an. Show that f

(x) 0. (See Example 2.12.)

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EXAMPLE 2.11. Show that the sequence an =n

n2 + 1is decreasing.

SOLUTION . We need to show that for every n, we have an an+1. This is the same as toshow that for every n, we have an an+1 0. We work this out:

an an+1 = nn2 + 1

n + 1(n + 1)2 + 1

=n(n + 1)2 + n (n + 1)(n2) (n + 1)

(n2 + 1)((n + 1)2 + 1)

=(n3 + 2n2 + n) (n3 + n2) 1

(n2 + 1)((n + 1)2 + 1)

=n2 + n 1

(n2 + 1)((n + 1)2 + 1) 0

EXAMPLE 2.12. Show that the sequence an =n

n2 + 1is decreasing.

SOLUTION . Let f(x) =x

x2 + 1. Then

f(x) =1 x2

(x2 + 1)2 0, for every x 1.

Thus f(x) is decreasing on the interval [1, ). Therefore, f(n + 1) f(n) for every n. Hencean+1 an for every n 1.

DEFINITION 2.13. A sequence {an} is said to be bounded above if there is anumber M such that an M for every n 1.It is said to be bounded below if there is a number m such that an m forevery n 1.

If it is bounded above and bounded below, then it is said to be bounded.

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Exercise 2.14. Can you think of a sequence which is bounded above but not bounded?Bounded below but not bounded? Neither bounded above nor bounded below?

THEOREM 2.15. If a sequence {an} is convergent, then it is bounded (both aboveand below).

THEOREM 2.16. Suppose that the sequence{an} is convergent, and let m an M for all n 1. Then,

m limnan M.

Note that if a sequence {an} is increasing, then it is automatically bounded below (bym = a1). Similarly if a sequence {an} is decreasing, then it is automatically bounded above(by M = a1).

THEOREM 2.17 (Monotone Sequence Theorem). If the sequence {an} is boundedand monotonic, then {an} is convergent.

There are different variations on the Monotone Sequence Theorem. For instance, we can saythat every increasing sequence which is bounded above is convergent. We can also saythat every decreasing sequence which is bounded below is convergent. The diagram belowillustrates this:

Note that the Monotone Sequence Theorem only allows us to prove that a given sequenceis convergent, but does not tell us what the actual limit value is. We normally have to workmore to calculate the actual limit.

Hence if we are given that a sequence {an} is increasing (or decreasing), to show that thissequence {an} is convergent, all we have to do is to find a bound M > 0 such that |an| < Mfor every n.

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Figure 2.2: Increasing and bounded above Figure 2.3: Decreasing and bounded below

The next theorem states that the convergence of a sequence is an eventual property ofa sequence. In other words, if two sequences are equal except for the first few terms, then thetwo sequences have the same limit.

THEOREM 2.18. The sequence {an} converges to L if and only if the sequence{an+M} converges to L, for any number M.

The following is an illustration with M = 11:

Figure 2.4: New sequence obtained by deleting the first 10 terms

In general, changing any finite number of a sequences terms does not affect the limit.

2.3 Limit Laws

The following are limit laws for sequences. They can help us to prove that certain sequencesconverge, and are also useful because they can help us compute limits.

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THEOREM 2.19 (Limit Laws for Sequences). If{an} and {bn} areconvergentandc is a constant, then

1. limn c = c, and limn can = c limn an.

2. limn(an bn) = limnan limn bn.

3. limn(anbn) = limnan limn bn.

4. limn

anbn

=limnan

limn bn

, provided that limn bn = 0.

5. limn(an)

p =

limn an

p, provided that p > 0 and an > 0 for all n.

You should take care when applying the limit laws. They can only be applied whenever {an}and {bn} are both convergent. If one or both of the sequences are divergent, then you cannotapply the limit laws.

EXAMPLE 2.20. We already know that limn

1

n2= 0. Using limit laws, we can conclude

that

limn

1

n3= lim

n

1

n23

2

= limn1

n23

2

= 0, let an =1

n2and p =

3

2and that

limn

3n2 + 4n3

n5

= lim

n

3n2

n5+

4n3

n5

= lim

n

3

n3+

4

n2

= 3 limn

1

n3+ 4 lim

n1

n2= 0

an = 3 1

n3and bn = 4 1

n2

One trick to find the limit of a fraction is to divide the numerator and the denominatorby the biggest occurring term.

EXAMPLE 2.21. Find limn

3n + 1

n2 1 .

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SOLUTION . We have3n2 + 2n

n2 1 =3 + 2

n

1 1n2

. Hence

limn

3 + 2n

1 1n2

=limn3 + 2n

limn

1 1

n2

= limn 3 + limn 2nlimn 1 limn

1

n2

=3 + 0

1 0 = 3

EXAMPLE 2.22. Find limn 2n +

n

1

n +

3n + 4

n +

n

SOLUTION . What is the largest term (ignoring multiplication by constants) in the numer-ator? For large n, 2n is much larger than

n 1 and so the largest / dominant term in the

numerator is

n.

What about the denominator?

n will get eaten up by n, so 4

n +

n becomes approxi-

mately 4

n. For large n this term is itself eaten up by 3n, and so

3n + 4

n +

n becomes

approximately 3n + 4

n which in turn becomes

3n =

3

n. So the dominant term in

the denominator is n.Since the dominant term in both the numerator and the denominator is

n, we divide the

numerator and the denominator by

n. We get2n +

n 1

n +

3n + 4

n +

n=

1n

2n +

n 1

1n

n +

3n + 4

n +

n

=

2 +

n1n

1 +

3 + 4

1n

+ 1nn

=

2 +

1n

1n2

1 +

3 + 4

1n

+ 1nn

The limit of this last expression is

2 +

0 0

1 +

3 + 4

0 + 0=

2

1 +

3.

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THEOREM 2.23. [Squeeze Theorem for Sequences] Suppose the sequences{an}, {bn} and {cn} satisfy

an bn cn for all large n > N,

and limn an = limn cn = L. Then limn bn must exist and is equal to the common

value L.

The idea is that suppose we are given a sequence {bn}, and we do not know if it is convergent.

If we are able to squeeze it between two sequences {an} and {cn} which we know have thesame limit, then we can also conclude that {bn} must have the same limit. The following is adiagram illustrating this:

Figure 2.5: The sequence {bn} is squeezed between {an} below and {cn} above.

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EXAMPLE 2.24. Determine whether the sequence an =cos n

nis convergent.

SOLUTION . If you are asked a question about a sequence involving trig functions, often youwould have to apply the Squeeze Theorem. In this case,

1n

cos nn

1n

[Since 1 cos n 1]

limn

1

n= 0

limn

1

n= lim

n1

n= 0 = 0 [Limit Law]

limn

cos n

n= 0 [Squeeze Theorem]

EXAMPLE 2.25. Show that if limn |an| = 0, then limnan = 0.

SOLUTION .

|an| an |an| [This is true for any number, whether positive or negative]limn|an| = limn |an| = 0 = 0

limn

an = 0 [Squeeze Theorem]

Warning: If the limn |an| = 1 (or any non-zero number), we cannot conclude that

limnan = 1. For instance, the sequence an = (1)

n. Then we have |an| 1 but an hasno limit.

Exercise 2.26. Show that limn |an| = 0 if and only if limnan = 0.

Exercise 2.27. Find limn

sin(en)

n. Find lim

n2n + sin(en)

n +

n.

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EXAMPLE 2.28. Find limn(

n + 1 n).

SOLUTION . Here we have another trick to simplify such an expression. Whenever we see

x y, we can try and rationalize by multiplying the expression by x + yx +

y

. Applying

this, we get

n + 1 n = (

n + 1 n)(n + 1 + n)

n + 1 +

n=

1n + 1 +

n

How do we find the limit of this last expression? We have

0 1n + 1 +

n

1n

and so by the Squeeze Theorem, limn(n + 1 n) = limn1

n + 1 + n = 0.

A quicker way to find limn

1n + 1 +

n

without having to apply the Squeeze Theorem

is to plug in . We can basically check quickly by substituting n with . In thiscase we get

1 + =1

= 0. Why cant we plug in into the original expression?We get which is an indeterminate form. You have to be very careful when youuse this method to find limits, as very often you will get an indeterminate form like

, 0,

,

0

0. If you get such an indeterminate form you have to simplify the ex-

pression further before trying to plug in again (just like what we did in the example above).If you are asked in a question to find the limit of a given sequence, or to show that asequence is convergent, you may apply the Limit Laws or the Squeeze Theorem. If you areexplicitly asked to prove by the definition, you have to use the definition of the limit, i.e. startwith and find N().

2.4 Other properties of limits

THEOREM 2.29. The geometric sequence {rn} is convergent if 1 < r 1, anddivergent for all other values of r. Specifically,

limn r

n =

0, if 1 < r < 1,1, if r = 1.

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DEFINITION 2.30 (Subsequences). Given a sequence {an}, a subsequence is thesequence {bn} remaining after deleting terms (possibly infinitely many) from {an}.

EXAMPLE 2.31. a5, a7, a8, a9, a10, . . . and a2, a4, a6, are both subsequences of {an}.However a2, a1, a3, a4, a5, is not a subsequence, as it contains terms in the wrong order.

EXAMPLE 2.32. Let an = (1)n. Taking bk = a2k and ck = a2k+1 we obtain two subse-quences. The first {bk}k=1 is called the even subsequence of {an}n=1. The second {ck}k=1 iscalled the odd subsequence of {an}n=1. Can you write out the terms of both subsequences?What are they?

EXAMPLE 2.33. Let an = ln n. Then {bk}k=1 is a subsequence of {an}n=1 where bk =ak3 = 3 ln k.

THEOREM 2.34. (i) If the sequence{an} converges to L then any of its subse-quence must converge to L.

(ii) If both the odd subsequence{a2n1} and the even subsequence{a2n} of a sequenceconverge to L then the sequence itself must also converge to L.

Theorem 2.34 is very useful, because it can be used to prove that a given sequence con-verges. To do this, we evaluate the limit of the even subsequence

{a2n

}= a2, a4, a6, a8,

and evaluate the limit of the odd subsequence {a2n1} = a1, a3, a5, a7, , and show that thetwo limits are equal. For example:

EXAMPLE 2.35. The sequence

(1)n

n

is convergent, because the even subsequence

converges to 0, and the odd subsequence converges to 0.

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Theorem 2.34 has another important use: We can use it to show that a given sequence isdivergent. To show that a given sequence {an} diverges, we can either (i) find a subsequence of{an} which has no limit, or (ii) find two subsequences of{an} which converge but to differentvalues. For example:

EXAMPLE 2.36. The sequence (1)n has no limit because the even subsequence convergesto 1 while the odd subsequence converges to 1.

2.5 Divergent sequences

We have seen the definition of what it means for a sequence to converge. A sequence is said

to diverge, if it does not converge to any L. However, there are special ways in which asequence can diverge. This is what we will study next:

DEFINITION 2.37. A sequence {an} is said to diverge to (or ) if for everyM > 0 there is a corresponding integer N(M) such that

whenever n > N(M) = an > M (or an < M).

In this case we write limnan = (or limnan = ).

Note that in this definition we are assuming that M is a very large positive real number. Inthe previous definition we are assuming that is a very small positive real number. Again,graphically we may visualize this in Figures 2.6 and 2.7.

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Figure 2.6: Pick N(M) depending on M.

Figure 2.7: Another visualization. N corresponds to M while N corresponds to M.

Again, note that N(M) depends on M. Generally, the larger the value of M, the largerthe value we have to choose for N(M). In Figure 2.6, what happens if we take M = 300? Canwe still pick N(300) to be the same?

Note that a sequence which diverges to

need not always be increasing, but generally

has to get arbitrarily large (see Example 2.38). Similarly, a sequence which diverges to need not always be decreasing.

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EXAMPLE 2.38. Show that the following sequence diverges to .

an =

n if n is odd,

n if n is even.

SOLUTION . The first few terms of the sequence are:

{1, 2,

3, 4,

5, 6,

7, 8, 3, 10,

11, 12, }

So this sequence is not increasing, but does generally grow larger and larger as n increases.Now we show that lim

n an = . Let M > 0 be given. We need to find the correspondingN(M). Again, we work backwards.

an > M =

n > M and n > M [Notice the one way implication]

n > M [Since n n, we make the smaller quantity > M] n > M2.

Now we have made n the subject. On the right we have M2. So we take N(M) =

M2

(again, we round up because we would like N to be an integer). Now to check that our choiceof N(M) works: We start with n > N(M) n > M2 n > M n > M and n >M an > M, following our working above. The one-way-implication above has to be thecorrect direction, otherwise, our final choice of N(M) will not be correct.

THEOREM 2.39. A sequence {an} can diverge only due to one of two reasons:(I) The sequence {an} contains two subsequences which converge to two different

limit values.

(II) The sequence {an} contains a subsequence that diverges to either or .

An example of a Type (I) divergent sequence:

{(1)n}The even subsequence a2, a4, a6, a8, converges to 1, while the odd subsequencea1, a3, a5, a7, converges to 1:

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Figure 2.8: Type I divergent sequence

An example of a Type (II) divergent sequence:an = n cos

n2

Notice that {an} has a subsequence diverging to , a subsequence converging to 0, and asubsequence diverging to . Can you pick them out?

Figure 2.9: Type II divergent sequence

EXAMPLE 2.40. Can you identify if the following geometric sequences are convergent,Type (I) divergent, or Type (II) divergent?

{0.75n

},

{(

0.75)n

},

{1.15n

}and

{(

1.15)n

}.

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Remember that the Limit Laws (Theorem 2.19) only works if both sequences converge. Ifone or both sequences diverge, we cannot apply the Limit Laws (Theorem 2.19). However inthat case we have a version of the Limit Laws for divergent sequences:

THEOREM 2.41 (Limit Laws for Divergent Sequences). Take sequences {an} and{bn}.

1. If{an} is divergent and c = 0 is a constant then {can} is divergent.2. If{an} is divergent and lim

n{bn} = L exists, then {an + bn} is divergent.

3. If{an} is divergent and limn{bn} = L = 0 exists, then {anbn} is divergent.

4. If{an} is divergent and limn{bn} = L = 0 exists, thenanbn is divergent.

PROOF . We use a proof technique called proof by contradiction. The way this works is asfollows. We start by assuming that the statement P we want to prove is false. We thenproceed to produce a proof of another statement Q which is contradictory. This Q can eitherbe something absurd, or it can be a contradiction to an earlier assumption. Since we havemanaged to produce a contradiction, the only logical conclusion is that our starting assumptionmust be incorrect, i.e. the statement P we want to prove cannot be false, i.e. the statementP must be true after all! (It is wierd, but it does work!)

(i): Suppose that {an} is divergent and c = 0. We wish to show that {can} is divergent.We start by assuming that the sequence {can} is not divergent, i.e. the sequence {can} isconvergent. Since c = 0, we can apply the Limit Law for Sequences (Theorem 2.19) to concludethat the sequence

1

c can

is convergent. That is, we conclude that {an} =

1

c can

is

convergent. This is a contradiction to our earlier assumption that {an} is divergent. Therefore,this contradiction we obtain leads us to conclude that the sequence {can} is divergent.

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Which is the statement P and which is Q?(ii): Suppose that {an} is divergent and lim

n{bn} = L exists. We wish to show that thesequence

{an + bn

}is divergent. We proceed to prove by contradiction. We start by assuming

that the sequence {an + bn} is convergent. By the Limit Law for Sequences (Theorem 2.19)we can conclude that the sequence {(an + bn) (bn)} is convergent. Hence we conclude thatthe sequence {an} = {(an + bn) (bn)} is convergent, which is a contradiction to our earlierassumption that {an} is divergent. This contradiction leads us to conclude that {an + bn} isdivergent.

Exercise 2.42. Try and prove parts (iii) and (iv) of Theorem 2.41 on your own.

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Chapter 3

Limit of a function

3.1 Definition

Consider a function f : A R. We have already defined the limit of a sequence. In a similarway we can define the limit of a function. There is one major difference between limits ofsequences and limit of functions: For limit of sequences we always consider n (n tendsto ). For limit of functions we consider x a (x tends to a or x approaches a).

Roughly speaking, we say that limxa f(x) = l if the value of f(x) can be made as close to l

as we like, by taking the value of x close to a (but not equal a).

DEFINITION 3.1. A function f approaches limit l as x approaches a point a

means:

For any > 0 there is () > 0 such that whenever x satisfies 0 0. Our task is find avalue of () that works. Atain we start with

|f(x) l| < |17x 34| < 17|x 2| < |x 2| <

17

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Now we have made |x a| = |x 2| the subject. Hence we can read off () = 17

. Note that

for function limits it is acceptable to take () to be a very small real number, and so we donot need to do any rounding off.

Now we verify that our choice of () = 17 works. We start with

0 < |x a| < () |x 2| < 17

|f(x) l| < .

EXAMPLE 3.3. Prove by the definition of the limit, that limx0

x sin 1x

= 0.

SOLUTION . Here we have f(x) = x sin1

x, l = 0 and a = 0. Given > 0, we need to make

|f(x) l| < x sin 1x

< .It seems hopeless to try and make |x a| = |x| the subject here. Fortunately we dont needto do this. We can try and replace

|f(x)

l

|= x sin

1

x by something larger and simpler:

x sin 1x = |x|

sin 1x |x|

which holds for every value ofx (more importantly this relationship must hold for every valueof x close to a = 0). Thus we can replace |f(x) l| < with |x| < , which is the same as|x a| < . So now we can read off () = .

So now we can check that our choice of () = works. We need to check by starting with0 < |x a| < ():

0 < |x a| < () |x| < Since |x| is larger than x sin 1x for every x close to a

x sin 1x

< |f(x) l| < .

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Exercise 3.4. Prove by the definition of the limit that limx3

2x + 2 = 8.

Exercise 3.5. Prove by the definition of the limit that limx1

x2 = 1.

Exercise 3.6. Prove by the definition of the limit that limx

1

1

x= 1.

Exercise 3.7. Prove by the definition of the limit that lim

x03

1 + x = 1.

THEOREM 3.8. Suppose the functionsf(x) andg(x) are equal for everyx in someopen interval containing the point x = a. If lim

xa f(x) = l we also have limxa g(x) = l.

Note that in the above theorem, f(a) and g(a) need not be equal.

3.2 Limit laws

We present the Limit Laws for function limits. These are similar to the Limit Laws forsequence limits (Theorem 2.19).

THEOREM 3.9. Given two functions f(x) and g(x), assume that limxa f(x) = l andlimxa g(x) = m, then

1. limxa(Af(x) Bg(x)) = A limxa f(x) B limxa g(x) = Al Bm.

2. limxa(f(x)g(x)) = limxa f(x) limxa g(x) = lm.

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3. limxa

f(x)

g(x)=

limxa f(x)

limxa g(x)

=l

m, provided m = 0.

4. limxa(f(x))

p =

limxa f(x)

p= lp provided f(x), l 0.

Again you should take care when applying the limit laws. They can only be applied wheneverlimxa f(x) and limxa g(x) both exist.

Exercise 3.10. Prove by definition of the limit that limxa x = a.

Exercise 3.11. What is limx0

x2

x? Can you apply the limit laws?

Exercise 3.12.

limxa

a0 + a1x + a2x2 + + anxn

= lim

xa a0 + limxa a1x + limxa a2x2 + + lim

xa anxn

= limxa

a0 + a1 limxa

x + a2 limxa

x2 +

+ an lim

xaxn

= limxa a0 + a1

limxa x

+ a2

limxa x

2+ + an

limxa x

n= a0 + a1a + a2a

2 + + anan

Can you identify which Limit Law we used at each step?

The above exercise shows that for polynomial functions p(x), we always have limxap(x) =

p(a). That is, to evaluate the limit for polynomials, we simply plug in the limit value for x.This is because polynimials are always continuous. Warning: You cannot always plug inthe limit value to find lim

xa f(x) for any function; you can only do this for functions which are

continuous! We will discuss continuous functions more in the next chapter.

EXAMPLE 3.13. Find limx2

x2 x 2x 2 .

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SOLUTION . Since we dont care about x = 2 when evaluating limits, we let g(x) =x2 x 2

x 2 and f(x) = x + 1. Then we see that

g(x) =x2 x 2

x 2 =(x + 1)(x 2)

x 2 = f(x)

for every x = 2. Applying Theorem 3.8 we see that limx2

g(x) = limx2

f(x) = 2 + 1 = 3. That is,

Theorem 3.8 allows us to simply a function g(x) algebraically before taking limits.


3

1 + x 1x

.

SOLUTION . Lets try and simplify the given expression before taking limits. Here is anothertrick we can use when faced with cube roots. We use the identity

(a b)(a2 + ab + b2) = a3 b3

Letting a = 3

1 + x and b = 1 we see that

a bx

=(a b)(a2 + ab + b2)

x(a2 + ab + b2)

=a3 b3

x(a2

+ ab + b2

)=

(1 + x) 13x

( 3

1 + x)2 + 3

1 + x + 1

=1

3

1 + x2

+ 3

1 + x + 1

So now (by Theorem 3.8) we see that

limx0

3

1 + x 1x

= limx0

13

1 + x2

+ 3

1 + x + 1=

1

3.

Similar to Theorem 2.23, we also have a version of the Squeeze Theorem for functions:

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THEOREM 3.15 (Squeeze Theorem for Functions). Given three functionsf(x), g(x) and h(x), assume that

f(x) g(x) h(x)

for every x I, x = a, where I is some open interval containing x = a. Then iflimxa f(x) = limxa h(x), we must also have limxa g(x).

Note that in order to apply the Squeeze Theorem for Functions, we only need to check thatf(x) g(x) h(x) for every x close to, but not including a.

Figure 3.2: g is squeezed between f and h.

Notice that in the above diagram the inequality f(x) g(x) h(x) holds only for all xclose to a (if x is far to the left then g(x) is larger than h(x)).

EXAMPLE 3.16.

Find limx0 x cos

1

x .

SOLUTION . Note that 1 cos 1x

1 for every x = 0. Thus we have

|x| x cos 1x

|x|

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for every x = 0. Since limx0

(|x|) = limx0

|x| = 0, we conclude by the Squeeze Theorem thatlimx0

x cos1

x= 0.

Notice that in this example, the function x cos 1x

is not defined at x = 0, but we can still

apply the Squeeze Theorem.


esin(cot x)x4.

SOLUTION . Note that sin(cot x) is always between 1 and 1, for all x close to but not equal0. Since ez is an increasing function, we get

1 sin(cot x) 1 = e1 esin(cotx) e1 = e1x4 esin(cotx)x4 ex4

Now limx0

ex4 = e limx0

x4 = 0 and limx0

e1x4 = e1 limx0

x4 = 0, hence by the Squeeze

Theorem, limx0

esin(cotx)x4 = 0

Fact 3.18. For 0 < x 0, we need to find (). Again we start with the end

|f(x) l| < |x2 0| < |x| < |x 0| <

In the line above we had two different choices to use as f(x). Why have we chosen x2 insteadofx +4 to substitute for f(x)? This is because we are interested in the right hand limit and sowe only consider values of x larger than a = 0; hence for these values of x we use f(x) = x2.

Now we take () = . To verify that our choice of works, we begin with taking xsatisfying:

a < x < a + () 0 < x < = |x| < following the statements above backwards |f(x) l| <

Notice that in the above, as we move from the first to the second line we have an implication= instead of . However this is okay because we only have to start from the statement

a < x < a + () and end with the statement |f(x) l| < .Can you check that the left-handed limit limx0 f(x) exists?Now lets check that lim

x1f(x) exists. Since x approaches 1, for all values of x close to

1, we get f(x) = x + 4. Thus the limit should be f(1) = 1 + 4 = 3. Lets use this as l.Given > 0, our task is to determine (). We start with the end

|f(x) l| < |(x + 4) 3| < |x (1)| < (3.1)

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Since a = 1, we have already made |x a| the subject. This suggests us to take () = .Do you think this will work? There is actually a little error in our statement 3.1 above. Sincewe are considering the regular two-handed limit, we need to consider all values of x close to

a = 1, including those x > 1. However if x > 0 then we cannot substitute x + 4 forf(x) (we must use f(x) = x2 instead for these values of x). Therefore in order that we canmake the substitution f(x) = x + 4 in 3.1 above, we need to choose () < 1, so that theinterval (a , a + ) = (1 , 1 + ) (2, 0). This extra restriction forces us to choose() = min{, 1}.

Lets try and verify our choice. Since this is the regular two-handed limit, we need to startby considering an x satisfying:

0 < |x a| < () = 0 < |x + 1| < and 0 < |x + 1| < 1= 0 < |x + 1| < and x < 0= |(x + 4) 3| < and x < 0

= |f(x) 3| <

THEOREM 3.31. Given a function f(x) and a number a. Then limxa f(x) = l if

and only if both one-handed limits exist and limxa+

f(x) = l and limxa

f(x) = l.

PROOF . We have to prove two directions. First suppose that limxa f(x) = l. This means that

we have a function 1() corresponding to limxa f(x) = l.

We must now show that limxa+

f(x) = l and limxa

f(x) = l. Lets first show that limxa+

f(x) =

l (the other case limxa

f(x) = l is similar). Given > 0, we must find the corresponding ().

We take()

We must find this value

= 1() The given corresponding to lim

xa

f(x) = l

Now it is easy to check that our choice for () works. Since we are proving a right-handedlimit, we begin by taking

a < x < a + () = a < x < a + 1()= x (a, a + 1)= x (a 1, a) (a, a + 1)= 0 < |x a| < 1= |f(x) l| <

By the definition of 1() and lim

xa f(x) = l

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The left handed limit is similar.Now we prove the opposite direction. We assume that lim

xa+f(x) = l and lim

xaf(x) = l.

This assumption means that we are given 1() and 2() corresponding to lim

xa+

f(x) = l and

limxa

f(x) = l respectively.

Now we are given > 0 and we must find () to satisfy limxa f(x) = l. We take () =

min{1(), 2()}. We now verify that this choice for () works. Since we are verify that works for the two-handed limit, we begin with

0 < |x a| < () = x (a (), a) or x (a, a + ())

Now if x (a (), a) then x (a 2(), a) because 2() (). Then by definition of2() and the left-handed limit, we conclude that |f(x) l| < .

The other case to consider is when x (a, a+()). Then we certainly have x (a, a+1())because 1()

(). By definition of 1() and the right-handed limit, we conclude that

|f(x) l| < .We have showed that in either case |f(x) l| < .

To summarize, there are roughly two ways we can proceed to prove that limxa f(x) does not

exist:

[Method 1] (Theorem 3.31) We can argue that one of the one-handed limits limxa

f(x)

or limxa+

f(x) does not exist, or we can argue that they both exist but are unequal.

[Method 2] (Theorem 3.22) We can either find a sequence an = a converging to asuch that limn f(an) is divergent, or find two sequences {an} and {bn} which are bothconverging to a and an = a, bn = a, such that lim

n f(an) and limn f(bn) are bothdifferent.

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Chapter 4

Continuity

4.1 Continuous functions

One of the main reasons to introduce the concept of limit is to define continuous functions,and further, the derivative and the integral. Naively one can say that a function is continuousif its graph is a solid line so it can be drawn without lifting the pen from the paper. Do youthink this is a good definition?

There are two issues: First, we need a more rigorous definition. Defining continuity interms of pens and papers is simply unacceptable! Second, the definition using drawing thegraph without lifting off the pen is good only when defining the continuity of a function onan interval. However in mathematics we often talk about the continuity of a function at asingle point. We need to start getting used to talking about continuity of a function f(x) ata single point a.

DEFINITION 4.1. A function f(x) is continuous at a point a R if

limxa f(x) = f(a).

We say that f(x) is continuous on the interval I if f(x) is continuous at each pointa I.

We have two kinds of continuity: continuity at a single point, and continuity on an interval.

To check that f(x) is continuous at a point a, we need to check that all three conditions hold:(i) limxa f(x) = L exists, (ii) f(a) exists, and (iii) L = f(a). If one or more of these conditions

are violated then we say that f(x) is discontinuous at a.

Lets look at Figure 4.1. In the first diagram on the left, the function f(x) is continuous atx = a, because lim

xa f(x) exists and is equal f(a). In the second middle diagram, the function

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Figure 4.1: Which function is continuous?

is discontinuous even though limxa f(x) exists, because the function f(a) is not defined atx = a. Finally in the third diagram on the right, the function is discontinuous even thoughboth lim

xa f(x) and f(a) exist, because the two values are different.

Figure 4.2: Where is the function continuous?

Exercise 4.2. In Figure 4.2, for each value of x = 0, 1, 2, 3, 4, determine if the function iscontinuous at x. If the function is discontinuous, give the reason why.

EXAMPLE 4.3. Any polynomial function is continuous on (, ). In Example 3.12 wevealready shown that lim

xap(x) = p(a) for every a (, ).

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EXAMPLE 4.4. Prove that sin x is continuous on R = (, ).

SOLUTION . We need to check that limxa sin x = sin a for every a R. However this is the

same as showing that

limxa sin x = sin a

limxa sin x

sin a = 0

limxa (sin x sin a) = 0

limxa |sin x sin a| = 0 [By Example 3.21] .

Now lets try and show this last line by applying the Squeeze Theorem.

0 |sin x sin a|

2cos x + a2

sin x a2

[Trigo identity] 2 1

x a2 [From Fact 3.18 we know that | sin z| |z|]

= |x a|.

Applying the Squeeze Theorem, we get limxa |sin x sin a| = 0, and so we conclude that sin x

is continuous on R.

We need more example of continuous functions. To help us get more, lets look at inversefunctions.

Recall! Given a function f(x), how do you find its inverse f1? We set y = f(x) andwrite x in terms of y, i.e. make x the subject. Then we have x = f1(y). Please note that

f1(x) = 1f(x)

, the latter being the reciprocal.

EXAMPLE 4.5.The inverse of f(x) = x

n

is f1

(x) =n

x. The inverse of g(x) = ex

isg1(x) = ln x. The inverse of sin(x) is (unsurprisingly!) sin1(x). What is the inverse of 3x3?

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Figure 4.3: The graph of f1 is obtained by reflecting the graph of f about the line y = x.

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THEOREM 4.6. If f(x) is continuous on (a, b) and f1(x) exists, then f1(x) isalso continuous on [f(a), f(b)].

EXAMPLE 4.7. This tells us immediately that f(x) = n

x is continuous on [0, ) andg(x) = sin1(x) is continuous on [1, 1].

THEOREM 4.8 (Limit Laws for continuous functions). Given two continuous func-

tions f(x) and g(x), the functions

cf(x) dg(x), f(x)g(x), f(x)g(x)

and f(g(x))

are also continuous whenever they are defined.

A very important application of this Theorem is used to show that every elementaryfunction is continuous on its natural domain: An elementary function is a function which isobtained from constants, variables, exponential, logarithm, trigonometric and inverse trigono-metric functions by performing addition, subtraction, multiplication, division, power and func-tion composition.

EXAMPLE 4.9. Any polynomial function, rational function, and even the function |x| iselementary. Can you explain why each of these functions is elementary?

Basically any function which can be defined by a single formula (using ingredients thatyou have seen before) are elementary. Can you think of any function which is not elementary?

THEOREM 4.10. Any elementary function is continuous whenever it is defined.

Whenever we are given an elementary function f(x) and we are asked to find limxa f(x), we

can simply plug in the value a. Iff(a) exists, then by Theorem 4.10 we can safely concludethat lim

xa f(x) also exists and must equal f(a). For instance, limx2(2x3+4x+5) = 16+8+5 = 29.

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What happens if we compute f(a) and find that it does not exist? Can we conclude fromthis that lim

xa f(x) does not exist? The answer is no!!

EXAMPLE 4.11. The function f(x) =sin x

xis elementary and has domain (, 0)(0, ).

So f(0) does not exist, but from Example 3.20 we know that limx0

sin x

x= 1.

So plug in the value of a only works to find the limit if the function is elementary and iff(a) exists. If f(a) does not exist then we have to determine lim

xa f(x) by other means. Oneway is to expand the components off(x) using Taylors expansion and taking the asymptoticvalue (o-notation). We will not cover this in this course. The second way is to apply LHopitalsrule, which we will cover in a later chapter.

4.2 Three theorems on continuous functions

We now present three important theorems about continuous functions. They are intuitivelyobvious by inspecting the graph of a continous function. However we will omit the proofs.

Figure 4.4: Try and draw a continuous function without crossing the horizontal line. Is itpossible?

The first theorem is called the Intermediate Value Theorem. The intuitive explanation is

in Figure 4.4. Suppose that f is a continuous function on the interval [a, b]. In Figure 4.4, thegraph of f is represented by a continuous curve from the starting point to the ending point.Can you draw the graph of the continuous function f without lifting your pen off the paper,such that you avoid crossing the horizontal line?

The answer is no! This is precisely what the IVT says; every such red line y = M mustbe in the range of f, i.e. there must be some c such that f(c) = M. First we state the basic

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form of the Intermediate Value Theorem for M = 0:

THEOREM 4.12 (Intermediate Value Theorem). Suppose that f is continuous on[a, b] and that f(a) 0 f(b) (or f(b) 0 f(a)). Then there is at least one pointc [a, b] such that f(c) = 0.

Why is this theorem called the Intermediate Value Theorem? This is because everyintermediate value M between f(a) and f(b) must be in the range of f, as the followingExample shows.

EXAMPLE 4.13. Suppose that f is continuous on [a, b] and that f(a) M f(b) (orf(b) M f(a)). Then there is at least one point c [a, b] such that f(c) = M.

SOLUTION . Let g(x) = f(x) M. Then g(a) = f(a) M 0 f(b) M = g(b) andso we can apply the IVT to g(x). We get some c [a, b] such that g(c) = 0. Hence we getf(c) M = 0 and so f(c) = M.

Figure 4.5: This intermediate value M shown between f(a) and f(b) is hit three times. Otherintermediate values are hit twice, but all intermediate values are hit at least once.

EXAMPLE 4.14. Prove that f(x) = 2x3 x 2 has at least one real root between x = 1and x = 2.

SOLUTION . f(x) is a polynomial, so it is continuous everywhere. Let a = 1, b = 2, we havef(a) = 1 and f(b) = 12. By the Intermediate Value Theorem, we get some c [1, 2] suchthat f(c) = 0. This number c is a root of f(x).

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Exercise 4.15. Prove that the line y = 2x + 1 intersects the curve y = ex for some value ofx [1, 2].

Exercise 4.16. Prove that if f(x) is continuous on [a, b] and the range of f is finite, then

f(x) is constant.

THEOREM 4.17 (Continuous functions are bounded). Assume that f(x) is con-tinuous on the interval [a, b]. Thenf(x) is bounded, i.e. there are constants m andM such that m f(x) M for every x [a, b].

Exercise 4.18. Can you replace the interval [a, b] in the above Theorem with (a, b)? That is,must every continuous function on (a, b) be bounded?

THEOREM 4.19 (Extreme Value Theorem). Assume that f(x) is continuous onthe interval [a, b]. Then the extremal values (i.e. maximum and minimum values) off(x) are in the range of f. That is, we can find some points xm, xM [a, b] such thatf(xm) f(x) f(xM) for every x [a, b].

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Figure 4.6: Extreme Value Theorem. The global maximum and minimum values are hit by f.

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CY1201 Calculus Lecture Set 1.pdf