Schinazi R.B. From Calculus to Analysis

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140 5 Continuity, Limits, and Differentiation

This completes the proof of the operations on continuous functions.

A polynomial is a function P(x)=a

n−1

+···+a

x +a

, where n

is a nonnegative integer, and a

,0≤i ≤n, are real numbers.

Example 5.4 A polynomial is continuous everywhere on the reals.

We ﬁrst take care of power functions. Let k be a natural number, and let f

(x) =

.LetP be the set of natural numbers k such that f

is continuous everywhere.

Let a be a real number, and a

be a sequence converging to a. Then, of course,

) = a

converges to f

(a) =a. Thus, 1 belongs to P . Assume that k belongs

to P . Then a

k+1

converges to a

a =a

k+1

by (ii). By induction, we see that

for every natural number k, the function f

is continuous everywhere.

Consider now the function g

(x) = b

. Then by (ii) we see that g

is contin-

uous for all naturals k and all reals b

. Finally, a polynomial P is always a sum

P = g

+···+g

. Using (i) and an induction proof, one can show that P

is continuous everywhere. The details are left as an exercise.

Example 5.5 Consider the function h deﬁned by h(x) = 1/x for all x = 0. Let

D = (−∞, 0) ∪ (0, ∞); h is deﬁned on D. We now show that h is continuous

everywhere on D.

Let a be in D.Leta

be in D and such that a

converges to a. Then by (iii)

we have that 1/a

converges to 1/a. That is, h(a

) converges to h(a), and h is

continuous at a.

Example 5.6 The function square root is continuous on [0, ∞).

First take a = 0. Let x

be in D and converging to 0. For any >0, there is N

such that |x

|<

for n ≥N . Since x

is positive, |x

|=x

, and since square root

is an increasing function on D,

0 ≤

√





=||=.

This proves that the square root function is continuous at 0.

Take now a>0. Assume that y

is in D and converges to a. For any >0, there

is N such that if n ≥N,wehave

−a|<

√

Observe now that



√

−

√



−a|

√

Since

√

a ≥

√

and the inverse function is decreasing on (0, ∞),wehave

√

≤

√

5.1 Continuity 141

Therefore,



√

−

√



−a|

√

≤

−a|

√

Hence, for n ≥N,wehave



√

−

√



≤

−a|

√



√

=.

This concludes the proof that the square root function is continuous at a.

Another important operation is the composition. Let f be deﬁned at a, and g be

deﬁned at f(a). We deﬁne the function g ◦f at a by

g ◦f(a)=g



f(a)



Continuity and composition of functions

Let f be deﬁned on D, and let g be deﬁned on the range C of f :

C =



f(x):x ∈D



Assume that f is continuous at a ∈D and that g is continuous at f(a)in C.

Then g ◦f is continuous at a.

To prove the claim above, take a sequence a

in D that converges to a. Since

f is continuous at a, f(a

) converges to f(a). Since g is continuous at f(a) and

f(a

) is a sequence in C that converges to f(a),wehavethatg(f(a

)) converges

to g(f (a)). This shows that g ◦f is continuous at a, and we are done.

Example 5.7 Let g be continuous on the reals. Let h be deﬁned by h(x) =g(x

Then h is continuous everywhere.

Let f be deﬁned on the reals by f(x)= x

.Wehaveh =g ◦f . Since g and f

are known to be continuous everywhere, so is h, and we are done.

Let a<bbe two reals. We denote the set of all reals x such that a ≤ x ≤ b by

[a,b].Theset[a,b] is called a closed and bounded interval. It is said to be closed

because the two bounds a and b are in the set. Intervals are the only sets of reals

with no holes. This property is crucial for the next result. See Sect. 5.4 for a formal

deﬁnition of intervals.

Intermediate Value Theorem

Assume that the function f is continuous at every point of the interval [a,b].

Let c be strictly between f(a) and f(b). There exists d in (a, b) such that

f(d)=c.

142 5 Continuity, Limits, and Differentiation

We start by proving a particular case of the intermediate value theorem (IVT).

Assume that c = 0 and that f(a)<0 <f(b). We need to show that the equation

f(x)= 0 has at least one solution. The proof that we now write is interesting be-

cause not only it proves the existence of a solution, but also it provides an algorithm

to ﬁnd the solution up to an arbitrarily small error. We construct inductively two

sequences a

and b

such that

f(a

) ≤0,f(b

) ≥0,a

≤a

n+1

≤b

n+1

≤b

, and b

−a

≤

−a

Let a

=a and b

=b. Compute f(

If f(

)<0, set a

and b

If f(

)>0, set b

and a

If f(

) =0, set a

Note that in all cases we have f(a

) ≤ 0, f(b

) ≥ 0, a

≤ a

≤ b

, and

−a

≤

−a

Assume now that sequences a

and b

are deﬁned up to k =n and that

n−1

≤a

≤b

n−1

and b

−a

≤

−a

Then,

If f(

)<0, set a

n+1

and b

n+1

If f(

)>0, set b

n+1

and a

n+1

If f(

) =0, set a

n+1

Since a

≤b

,wehavethata

≤a

n+1

≤b

n+1

≤b

. Moreover, in all three cases

we have

n+1

−a

n+1

≤

−a

≤

−a

n+1

By induction we have constructed sequences a

and b

with the prescribed proper-

ties. Since

0 ≤b

−a

≤

−a

by the squeezing principle b

−a

converges to 0 (why?). We also have that a

an increasing sequence bounded above by b

(why?), so it must converge. Similarly,

is a decreasing sequence bounded below by a

, so it converges as well. More-

over, the two limits must be the same, and we denote the limit by d. Note that by

construction

f(a

) ≤0 and f(b

) ≥0.

Letting n go to inﬁnity in the two inequalities above and using that f is continuous

at d (why?), we get

f(d)≤0 and f(d)≥0.

5.1 Continuity 143

Thus, f(d)=0, and we have proved the intermediate value theorem in the particular

case c =0.

We now turn to the general case. Let c be strictly between f(a) and f(b).As-

sume that f(a)<c<f(b); the other case is treated similarly. Let the function g be

deﬁned by

g(x) =f(x)−c.

Note that g is also continuous on [a,b].Wehaveg(a) < 0 and g(b) > 0. According

to the particular case of the IVT that we already proved, there is d in (a, b) such that

g(d) =0. This implies that f(d)=c, and the IVT is proved.

Example 5.8 We use the algorithm above to ﬁnd an approximation of

√

2. Deﬁne

f(x)= x

. Note that f(1) = 1 and f(2) = 4 and that f is continuous on [1, 2]

(it is continuous everywhere of course, so also on [1, 2]). So by the IVT there is d

in (1, 2) such that f(d)= 2. That is, d

= 2 and d =

√

2. Let a

= 1 and b

= 2.

We have f(3/2) = 9/4 > 2, so a

= 1 and b

= 3/2. The midpoint of 1 and 3/2

is 5/4, and f(5/4) = 25/16 < 2. Thus, a

= 5/4 and b

= 3/2. That is,

√

2isin

). An approximate value of

√

2 is then (a

+ b

)/2 = 11/8.Theerrorisat

most (b

−a

)/2 =(b

−a

)/2

=1/16.

Example 5.9 Let g be deﬁned by

g(x) =−1forx ∈[−1, 0) and g(x) =1forx ∈(0, 1].

The IVT does not apply: observe, for instance, that 0 is between −1 and 1, but that

there is no d such that g(d) = 0. The problem is that g is deﬁned and continuous

on [−1, 0) ∪(0, 1] which is not an interval! It has a hole at 0. This is why the IVT

cannot be applied.

We now turn to another important theorem concerning continuous functions.

Extreme Value Theorem

Assume that the function f is continuous at every point of the interval [a,b].

Then f attains both a minimum and a maximum value. That is, there are real

numbers c and d in [a,b] such that for all x in [a,b],

f(c)≤f(x)≤f(d).

We prove this theorem in two steps. In the ﬁrst step we prove that a continuous

function on [a,b] must be bounded. That is, we want to show that there is M such

that for all x in [a,b],



f(x)



<M.

We do a proof by contradiction. Assume that for every natural n, there is x

in [a,b]

such that



f(x

)



>n.

144 5 Continuity, Limits, and Differentiation

Since x

is always in [a,b], the Bolzano–Weierstrass theorem applies, and there is a

subsequence x

that converges to some . It is easy to see that  is in [a,b] (why?).

Since f is continuous at ,sois|f | (why?). Hence, |f(x

)| converges to |f()|.

On the other hand,



f(x

)



That is, |f(x

)| is unbounded. This is a contradiction, a convergent sequence must

be bounded. Therefore, f is bounded.

In our second step we show that f attains a maximum (that f attains a minimum

is similar and left as an exercise). Consider

A =



f(x):x ∈[a,b]



That is, A is the range of f . Note that by the ﬁrst step, A is bounded above by M.

Clearly, A is not empty (why?). By the fundamental property of the reals, A has a

least upper bound m. As shown in Sect. 2.1, there is a sequence y

in A that con-

verges to m. Since y

is in A, there is z

in [a,b] such that y

=f(z

). Hence, there

is a sequence z

in [a,b] such that f(z

) converges to m. By Bolzano–Weierstrass

(since z

is bounded), there is a subsequence z

that converges to some d in [a,b].

Since f is continuous at d, f(z

) converges to f(d).Butf(z

) converges to m as

well since f(z

) is a subsequence of f(z

). Hence, we must have f(d)=m.This

proves that f attains its maximum at d.

Example 5.10 The extreme value theorem applies to continuous functions on closed

and bounded intervals. That is, intervals of the type [a,b]. Consider f(x)=1/x for

x on (0, 1]. The function f is continuous on (0, 1]. However, it is not bounded. For

every n,



1/(n +1)



=n +1 >n.

Hence, f cannot attain its maximum. The EVT does not apply since (0, 1] is not

closed at 0.

For most of our applications, we will use the deﬁnition of continuity using se-

quences. However, the following deﬁnition of continuity is also useful.

Another deﬁnition of continuity

Let f be deﬁned on D. Then, f is continuous at a ∈ D if and only if for

every >0, there is a δ>0 such that if x is in D and |x − a| <δ, then

|f(x)−f(a)|<.

We want to show that the statement A: ‘f is continuous at a’ is equivalent to the

statement B: ‘for every >0, there is a δ>0 such that if x is in D and |x −a|<δ,

then |f(x)−f(a)|<’.

5.1 Continuity 145

Assume non B. That is, there is >0 such that for all δ>0, there is x in D such

that |x − a| <δ and |f(x)−f(a)|≥. Since this must be true for all δ>0, we

may pick δ =1/n. There is x in D such that

|x −a|< 1/n and



f(x)−f(a)



≥.

In fact x is found once δ = 1/n is ﬁxed, therefore x depends on n, and we will

denote it by x

. Since we may do this for any δ =1/n, we get a sequence x

in D

with

−a|< 1/n for every n ≥1.

In particular, x

converges to a. On the other hand,



f(x

) −f(a)



≥.

Hence, f(x

) does not converge to f(a), and f is not continuous at a. This shows

that non B implies non A. This is logically equivalent to A implies B.

Assume B. Fix >0. There is δ>0 such that if D is in I and |x −a|<δ, then

|f(x)−f(a)|<.Leta

be a sequence in D that converges to a. There is N such

that if n ≥N , then

−a|<δ.

By B this implies



f(a

) −f(a)



< for n ≥N.

That is, f(a

) converges to f(a). The function f is continuous at a.Wehaveshown

that B implies A. This completes the proof that the two continuity deﬁnitions are

equivalent.

Application 5.1 Assume that g is deﬁned on D and continuous at a ∈D. Suppose

that g(a) > 0. Then, there exists δ>0 such that if x is in D and if |x −a|<δ, then

g(x) > g(a)/2.

We apply the second continuity deﬁnition with  = g(a)/2 > 0. There is δ>0

such that if x is in D and |x −a|<δ, then



g(x) −g(a)



<=g(a)/2.

In particular,

g(x) −g(a) > −g(a)/2.

That is,

g(x) > g(a)/2

for all x in D such that |x −a|<δ.

146 5 Continuity, Limits, and Differentiation

Exercises

1. We complete the proof that a polynomial is continuous everywhere in Exam-

ple 5.2. Show that for any natural n and any continuous functions g

,...,g

we have that g

+···+g

is continuous.

2. Show that g(x) =

1+x

is continuous everywhere.

3. Prove that a rational function (that is, the ratio of two polynomials) is continu-

ous where it is deﬁned.

4. Show that the number d deﬁned in the IVT is strictly between a and b.

5. Is the number d deﬁned in the IVT unique? Prove it or give a counterexample.

6. (a) Show that the equation x

+x +5 =0 has at least one solution.

(b) Find an approximation of a solution good to the ﬁrst decimal.

7. Assume that f is continuous on [−2, 2] and that f(−2)>−2, f(2)<2. Show

that the equation f(x)=x has at least one solution.

8. Let f and g be continuous on [0,1]. Assume that f(0)<g(0) and f(1)>g(1).

Show that the equation f(x)=g(x) has at least one solution.

9. Consider the function f(x)=

1+x

on R.

(a) Show that f attains its maximum.

(b) Does f have a minimum?

10. Complete the proof of the EVT. That is, assume that f is continuous on [a,b]

and show that f attains its minimum. (Recall that we proved that f is bounded

below and above.)

11. Let f be deﬁned on the interval I and continuous at a ∈I .

(a) Show that |f | is continuous at a.

(b) Show that f

is continuous at a.

12. Assume that g is deﬁned on D and continuous at a ∈D. Suppose that g(a) < 0.

Show that there exists δ>0 such that if x is in D and if |x −a|<δ, then

g(x) < 3g(a)/4.

13. Assume that g is a continuous function on D.Leta be in D and such that

g(a) > 0. Show that if a

is a sequence in D that converges to a, then there is

N such that if n ≥N , then g(a

)>0.

14. (a) Let f be a continuous function on R. Assume that f(r)=0 for every ratio-

nal r. Show that f(x)=0 for every real x. (Use the density of the rationals.)

(b) Let f and g be continuous functions on R. Assume that f(r)= g(r) for

every rational r. Show that f(x)=g(x) for every real x.

15. Let f be deﬁned on R and suppose that there is a real M>0 such that



f(x)−f(y)



≤M|x −y|

for all x and y. Show that f is continuous on R.

16. Let f be deﬁned on [0, ∞) by

f(x)=min



x,x



(a) Sketch the graph of f .

5.2 Limits of Functions and Derivatives 147

(b) Show that for all x and y,

min(x, y) =



(x +y) −|x −y|



Show that h is continuous on D.

17. Let a be a real. Deﬁne the function f by f(x)=|x −a|. Prove that f is con-

tinuous on R.

18. Let f be a continuous function on [0, 1]. Assume that f ≥0.

(a) Show that either there is c such that f(c)= 0 or there is a>0 such that

f(x)≥a for all x in [0, 1] (in the latter case, f is said to be bounded away

from 0).

(b) Give an example of a function g ≥ 0 continuous on (0, 1] for which (a)

does not hold.

19. Deﬁne the function f as follows. If x is a rational, f(x)=

1. If x is not rational,

x) = 0. Show that f is nowhere continuous. That is, for any a in R,the

function f is not continuous at a. (Use that the rational numbers are dense

in the real numbers and that the irrational numbers are also dense in the real

numbers.)

5.2 Limits of Functions and Derivatives

We start with a deﬁnition.

Limit of a function

Assume that f is deﬁned on a set D ⊂ R. The function f is said to have a

limit  at a if for every sequence x

in D such that x

=a for all n ≥1 and

such that x

converges to a,wehave

lim

n→∞

f(x

) =.

The limit of the function f at a is denoted by

lim

x→a

f(x)=.

Note that in order for the deﬁnition above to make sense, the point a and the set

D must be such that there are sequences in D with no term equal to a and con-

verging to a. For instance, if D ={1, 2} and a =1, then there is no such sequence

(why not?). This is why one usually deﬁnes the notion of limit point of a set before

deﬁning limits of functions, see the Exercises. Since we will be interested only in

sets D that are intervals or union of intervals and in points a that are either in D or

at the boundary of D, we omit the notion of limit point for the time being.

148 5 Continuity, Limits, and Differentiation

Example 5.11 Let f be deﬁned on D =R by f(x)=x

.Leta =1. Does f have a

limit at 1?

Let x

be a sequence in R that converges to 1 and such that x

=1 for all n ≥1.

Then x

converges to 1 (why?). Hence, f(x

) converges to 1. This proves that the

limit of f at 1 exists and is 1.

Example 5.12 Let g be deﬁned on D =(−∞, 0) ∪(0, ∞) by

g(x) =

|x|

Does g have a limit at 0?

Intuitively, the answer is clearly no. The function g is −1forx<0 and 1 for

x>0. Depending on which side we approach 0, we get one limit or the other. We

now prove this. Let x

=1/n. Then x

is always in D, converges to 0, and is never 0.

Since x

> 0, we have g(x

) =1 for all n ≥1. Hence, g(x

) converges to 1.

Now, let y

=−1/n. Then y

is always in D, converges to 0, and is never 0.

Since y

< 0, we have g(y

) =−1 for all n ≥1. Hence, g(y

) converges to −1.

That is, we have two sequences converging to 0 but such that g(x

) and g(y

)

converge to different limits. Hence, the function g has no limit at 0.

Note that we picked two of our favorite sequences to prove that g has no limit at

a = 0. But if we wanted to prove that g has a limit, we would need to work with a

generic sequence x

converging to a as was done in Example 5.11.

Example 5.13 Let h be deﬁned on D =(0, ∞) by

h(x) =

√

Does h have a limit at 0?

By computing h(x) for a few x near 0 (0.1, 0.01,...) it is easy to see that h grows

without bound near 0 and therefore has no limit. We now prove it. Let x

= 1/n.

Then x

is always in D, converges to 0, and is never 0. We have h(x

) =

√

n.This

sequence is unbounded and therefore does not converge. The function h has no limit

at 0.

Remark Examples 5.12 and 5.13 give useful methods to prove that a function has no

limit. Assume that we want to show that function g has no limit at a. One method,

used in Example 5.12, is to ﬁnd two sequences a

and b

in the domain of g converg-

ing to a (but never equal to a) and such that g(a

) and g(b

) converge to different

limits.

The other method, used in Example 5.13, is to ﬁnd a sequence a

in the domain

of g, converging to a and such that g(a

) does not converge.

The following alternative deﬁnition of a limit is sometimes useful.

5.2 Limits of Functions and Derivatives 149

Alternative deﬁnition of limit

The function f deﬁned on D has a limit  at a if and only if for every >0,

there is a δ>0 such that if x is in D and 0 < |x −a|<δ, then |f(x)−| <.

Observe that this –δ deﬁnition of a limit is almost the same as the corresponding

deﬁnition of continuity at the send of Sect. 5.1. The only differences are that |x −

a|> 0 (which is the same as x =a) and the use of  instead of f(a). The proof that

this second deﬁnition is equivalent to the ﬁrst one is almost identical to the proof

that the two deﬁnitions of continuity are equivalent. We leave it as an exercise.

There is a close relation between the notion of limit and the notion of continuity.

Continuity and limit

Assume that the function f is deﬁned on D.Leta be in D. The function f is

continuous at a if and only if

lim

x→a

f(x)=f(a).

Assume ﬁrst that f is continuous at a. Take a sequence x

in D such that for all

n ≥1, x

=a and x

converges to a. By the deﬁnition of continuity at a we have

lim

n→∞

f(x

) =f(a).

This proves that f has a limit f(a)as x approaches a. This proves the direct impli-

cation.

For the converse, assume that the limit of f at a exists and is equal to f(a).

Use the alternative deﬁnition of limit to get: for every >0, there is a δ>0 such

that if x is in D and 0 < |x −a| <δ, then |f(x)− f(a)| <. Observe now that

|f(x)−f(a)|< holds for x = a as well. Hence, it holds for all x in D such that

|x −a|<δ. According to the second deﬁnition of continuity in Sect. 5.1, this proves

that f is continuous at a.

Example 5.14 Consider a polynomial p(x) =a

n−1

+···+a

x + a

where n is natural, and a

,...,a

are reals. For any real b, we have that

lim

x→b

p(x) =p(b).

Since a polynomial is continuous everywhere, we know that the limit of p at b is

p(b), and we are done.

Example 5.15 Consider the function

f(x)=

−4

x +2

for x =−2