MTAEA Continuity and Limits of Functions · Continuity and Limits of Functions Continuous Functions. Example Let f : R !R be given by f(x) = 2x2 +1 for all x 2R. Prove f is continuous

Continuity and Limits of Functions

MTAEA – Continuity and Limits of Functions

Scott McCracken

School of Economics,Australian National University

February 1, 2010

Scott McCracken MTAEA – Continuity and Limits of Functions


Continuous Functions.

I A continuous function is one we can draw without taking our penoff the paper

Definition.Let f be a real-valued function whose domain is a subset of R.

I The function f is continuous at x0 in dom(f ) if, for every sequence(xn) in dom(f ) converging to x0, we have lim f (xn) = f (x0).

I If f is continuous at each point of a set S ⊆ dom(f ), then f is saidto be continuous on S.

I The function f is said to be continuous if it is continuous ondom(f ). N




x

y

x0

Figure: A continuous function. For any sequence of points (xn)n∈N converging to x0, thesequence (f (xn))n∈N converges to f (x0).




ExampleLet f : R→ R be given by f (x) = 2x2 + 1 for all x ∈ R. Prove f iscontinuous on R.

Proof.Suppose we have a real-valued sequence (xn) converging to x0 i.e.lim xn = x0. Then we have

lim f (xn) = lim[2x2n + 1] = 2[lim x2

n ] + 1 = 2x20 + 1 = f (x0),

where the second equality follows by application of the limittheorems. We have shown that for any sequence (xn) converging tox0, we have lim f (xn) = f (x0).This proves that f is continuous at eachx0 ∈ R and so f is continuous on R. �




ExampleLet f : R→ R be given by f (x) = (1/x) sin(1/x2) for x 6= 0 andf (0) = 0. Show that f is discontinuous at 0.

Proof.It is sufficient to find a sequence (xn) converging to 0 such that(f (xn)) does not converge to f (0) = 0. To find such a sequence, wewill rearrange (1/xn) sin(1/x2

n ) = 1/xn where xn → 0. Thus, we wantsin(1/x2

n ) = 1. For this we need 1/x2n = 2nπ + π/2, and it follows that

xn =1√

2nπ + π2.

Then lim xn = 0, while lim f (xn) = lim(1/xn) = +∞ 6= 0. �



Continuous Functions.I The sequential definition of continuity implies that the values f (x)

are close to f (x0) when the values x are close to x0. Thefollowing theorem provides an alternative definition of continuity.

Theorem (ε-δ definition of continuity)Let f be a real-valued function whose domain is a subset of R. Then fis continuous at x0 ∈ dom(f ) iff

for each ε > 0 there exists δ > 0 such thatx ∈ dom(f ) and |x−x0| < δ imply |f (x)−f (x0)| < ε.

I If we draw two horizontal lines, no matter how close together, wecan always cut off a vertical strip of the plane by two verticallines in such a way that all that part of the curve which iscontained in the strip lies between the two horizontal lines.



Continuous Functions.Epsilon Delta Applet

x

y

f (x0)− εf (x0)

f (x0) + ε

x0 − δ x0 x0 + δFigure: ε-δ definition of continuity


http://www.slu.edu/classes/maymk/Applets/EpsilonDelta.html



ExampleLet f : R→ R be given by f (x) = x2 sin(1/x) for all x 6= 0 andf (0) = 0. Prove f is continuous at 0.

Proof.Let ε > 0. Clearly |f (x)− f (0)| = |f (x)| ≤ x2 for all x . We want this tobe less than ε, so set δ =

√ε. Then |x − 0| < δ implies x2 < δ2 = ε,

and so|x − 0| < δ implies |f (x)− f (0)| < ε.

So f is continuous at 0. �




ExampleLet f : R→ R be given by f (x) = 2x2 + 1 for all x ∈ R. Then f iscontinuous on R.

Proof.We now use the ε-δ property. Let x0 ∈ R, and let ε > 0. We want toshow that |f (x)− f (x0)| < ε provided |x − x0| is sufficiently small , i.e.less than some δ. First note that

|f (x)− f (x0)| = |2x2 + 1− (2x20 + 1)| = |2x2 − 2x2

0 |= |2(x − x0)(x + x0)| = 2|x − x0||x + x0|.

So, what we need is to find a bound for |x + x0| that does not dependon x .




Now, if |x − x0| < 1 then |x | < |x0|+ 1 and hence|x + x0| ≤ |x |+ |x0| < 2|x0|+ 1. So, we have

|f (x)− f (x0)| < 2|x − x0|(2|x0 + 1)

if |x − x0| < 1. To arrange for 2|x − x0|(2|x0 + 1) < ε it is sufficient tohave |x − x0| < ε/[2(2|x0 + 1)] and also |x − x0| < 1. So let

δ = min{

1,ε

2(2|x0|+ 1)

}The working above shows that |x − x0| < δ implies|f (x)− f (x0)| < ε. �




I We can form new functions from old functions in several ways.

Definition.Let A ⊆ R and let B ⊆ R. Consider two functions f : A→ R andg : B → R and let k ∈ R. We define the functions into R as follows.

I |f | given by |f |(x) = |f (x)| for all x ∈ A;I kf given by (kf )(x) = kf (x) for all x ∈ A;I f + g given by (f + g)(x) = f (x) + g(x) for all x ∈ A ∩ B;I fg given by (fg)(x) = f (x)g(x) for all x ∈ A ∩ B;I f/g given by (f/g)(x) = f (x)/g(x) for all x ∈ A ∩ B such that

g(x) 6= 0. N




TheoremLet f and g be real-valued functions that are continuous at x0 in Rand let k ∈ R. Then

(i) |f | is continuous at x0;(ii) kf is continuous at x0;

(iii) f + g is continuous at x0;(iv) fg is continuous at x0;(v) f/g is continuous at x0 if g(x0) 6= 0.

TheoremIf f is continuous at x0 and g is continuous at f (x0), then thecomposite function g ◦ f is continuous at x0.



Properties of Continuous Functions.

Theorem (Intermediate Value Theorem)If f is a continuous real-valued function on an interval I, then f hasthe intermediate value property on I: Whenever a,b ∈ I, a < b and ylies between f (a) and f (b), there exists at least one x ∈ (a,b) suchthat f (x) = y.

I This theorem can be used to establish that a continuous functionf has a fixed point, i.e. a point x0 ∈ dom(f ) such that f (x0) = x0.




x

y0

f (b)

y0

f (a)

x1 x2 x3a b[ ]

I

f (x)

Figure: By the intermediate value theorem, for any f (a) < y < f (b), we can find an x such thatf (x) = y . In the case of y0, we can find three i.e. y0 = f (x1) = f (x2) = f (x3).



Properties of Continuous Functions.ExampleShow that the continuous function f : [0,1]→ [0,1] has a fixed pointx0 ∈ [0,1].

x

y

1

1

y=

x

f (x)

x0

Figure: Fixed point. Kinda obvious?




I We will use the IVT. We will define a new function g on I, andconsider a = 0 and b = 1.

Solution.Consider the function g(x) = f (x)− x , which is also continuous on[0,1] by (iii). Now

g(0) = f (0)− 0 = f (0) ≥ 0

andg(1) = f (1)− 1 ≤ 1− 1 = 0.

So, by the intermediate value theorem, we have that g(x0) = 0 forsome x0 ∈ [0,1]. Then clearly f (x0) = x0. �



Limits of Functions.

Definition.Let S be a subset of R, let a be a real number or symbol∞ or −∞that is the limit of some sequence in S, and let L be a real number orsymbol∞ or −∞. We write limx→aS f (x) = L if

I f is a function defined on S, andI for every sequence (xn) in S with limit a, we have

limn→∞ f (xn) = L. N

I The expression “limx→aS f (x)” is read “limit, as x tends to a alongS, of f (x)”.

I Using this definition, we see that a function f is continuous at a indom(f ) = S iff limx→aS f (x) = f (a).



Limits of Functions.Definition.For a function f and a ∈ R we write

(i) limx→a f (x) = L provided limx→aS f (x) = L for some set S = J\awhere J is an open interval containing a. limx→a f (x) is calledthe (two-sided) limit of f at a.

(ii) limx→a+ f (x) = L provided limx→aS f (x) = L for some openinterval S = (a,b). limx→a+ f (x) is called the right-hand limit of fat a.

(iii) limx→a− f (x) = L provided limx→aS f (x) = L for some openinterval S = (c,a). limx→a− f (x) is called the left-hand limit of fat a.

(iv) limx→∞ f (x) = L provided limx→∞S f (x) = L for some intervalS = (c,∞).

(v) limx→−∞ f (x) = L provided limx→−∞S f (x) = L for some intervalS = (−∞,b). N




x

y

t

s

a

f (x)

Figure: The right hand limit at a is t , while the left hand limit is s.



Limits of Functions.Example

(1) We have limx→4 x3 = 64 and limx→2(1/x) = 1/2 because thefunctions x3 and 1/x are continuous at 4 and 2 respectively. Onecan easily show that limx→0+(1/x) = +∞ andlimx→0−(1/x) = −∞. It follows that limx→0(1/x) does not exist(see theorem 7).

(2) Consider

limx→2

[x2 − 4x − 2

].

The function we are finding the limit of is not defined at x = 2. Wecan rewrite the function as

x2 − 4x − 2

=(x − 2)(x + 2)

x − 2= x + 2 for x 6= 2.



Limits of Functions.Now we can see that

limx→2

[x2 − 4x − 2

]= lim

x→2(x + 2) = 4.

It is important to note that the functions given by (x2 − 4)/(x − 2)and (x + 2) are not identical. The domain of the first is(−∞,2) ∪ (2,+∞), while the domain of the second is R.

(3) Consider

limx→1

[√x − 1

x − 1

].

To find the limit, we multiply the numerator and denominator ofthe function by

√x + 1 to get

√x − 1

x − 1=

x − 1(x − 1)(

√x + 1)

=1√

x + 1for x 6= 1.




Now we can see that

limx→1

[√x − 1

x − 1

]= lim

x→1

[1√

x + 1

]=

12.

This in fact shows that if h(x) =√

x then h′(1) = 1/2 (as you willsee when we look at differentiation).

(4) Let f be a real-valued function given by f (x) = 1/(x − 2)3 for allx 6= 2. Then

I limx→+∞ f (x) = 0,I limx→−∞ f (x) = 0,

I limx→2+ f (x) = +∞,I limx→2− f (x) = −∞.




Proof.We will show that limx→∞ f (x) = 0. By definition, it is enough to showthat limx→∞S f (x) = 0 for S = (2,+∞). So consider any sequence(xn) with xn ∈ S for all n ∈ N, such that limn→+∞ xn = +∞ and showthat

limn→+∞

f (xn) = limn→+∞

[1

(xn − 2)3

]= 0.

To prove the above assertion we could use our limit theorems.Instead we will prove it directly.Let ε > 0. We need to find an N such that for n > N, we have|(xn − 2)−3| < ε. This inequality can be rearranged asε−1 < |(xn − 2)3| or ε−

13 < |xn − 2|. We need xn > ε−

13 + 2, for the

previous inequality to be satsified.




Now we use the fact that limn→+∞ xn = +∞. By definition of aninfinite limit of a sequence, for any M > 0 we can find an N such thatn > N implies xn > M. Thus, if we set M = ε−

13 + 2, there exists an N

such thatn > N implies xn > ε−

13 + 2

Then reversing the steps above, we find that

n > N implies |(xn − 2)−3| < ε.

This shows that limn→+∞ f (xn) = 0 for any sequence (xn) in S suchthat limn→+∞ xn = +∞ and the result follows by definition. �




I The following result allows us to avoid sequences and provideε-δ definitions of a function’s limits.

TheoremLet f be a function defined on a subset S of R, let a be a real numberthat is the limit of some sequence in S, and let L be a real number.Then limx→as f (x) = L iff

for each ε > 0 there exists δ > 0 such thatx ∈ S, |x − a| < δ imply |f (x)− L| < ε.

I This theorem has a number of corollaries, which are listed in thenext theorem. These give us alternative definitions for the limit ofa function, its lateral limits, and its limits at infinity.



Limits of Functions.Theorem

(i) Let f be a function defined on J\{a} for some open interval Jcontaining a, and let L be a real number. Then limx→a f (x) = L iff

for each ε > 0 there exists δ > 0 such that0 < |x − a| < δ implies |f (x)− L| < ε.

(ii) Let f be a function defined on some interval (a,b), and let L be areal number. Then limx→a+ f (x) = L iff

for each ε > 0 there exists δ > 0 such thata < x < a + δ implies |f (x)− L| < ε.

(iii) Let f be a function defined on some interval (c,a), and let L be areal number. Then limx→a− f (x) = L iff

for each ε > 0 there exists δ > 0 such thata− δ < x < a implies |f (x)− L| < ε.




x

y

t

s + ε

ss − ε

a + δaa− δ

f (x)

Figure: The (two-sided) limit at a is s, while f (a) = t . (This function is not continuous at a.)




x

y

t + ε

tt − ε

s + ε

ss − ε

a + δaa− δ

f (x)

Figure: The right hand limit at a is t , while the left hand limit is s.




(iv) Let f be a function defined on some interval (c,+∞), and let Lbe a real number. Then limx→+∞ f (x) = L iff

for each ε > 0 there exists δ < +∞ such thatx > δ implies |f (x)− L| < ε.

(v) Let f be a function defined on some interval (−∞,b), and let Lbe a real number. Then limx→−∞ f (x) = L iff

for each ε > 0 there exists δ > −∞ such thatx < δ implies |f (x)− L| < ε.




x

y

f (x)

ε

−ε

1ε

−1ε

Figure: The function f : R \ {0} → R is given by f (x) = 1/x for all x ∈ R \ {0}. The limits at+∞ and −∞ are both 0.




ExampleLet f : R \ {0} → R be given by f (x) = 1/x for all x ∈ R \ {0} (this isthe function drawn in the previous figure). Thenlimx→+∞ = limx→−∞ = 0.

Proof.We will prove the first limit. First note that the function is defined onsome interval (c,+∞) – we could take any c > 0. Also the limit L = 0is a real number. So we use definition iv. Let ε > 0. We need to find afinite δ such that x > δ implies |1/x | < ε. If we take δ = 1/ε, then thedesired condition is satisfied. �



Limits of Functions.(vi) Let f be a function defined on J \ {a}. Then limx→a f (x) =∞ iff

for each M > 0 there exists δ > 0 such that0 < |x − a| < δ implies f (x) > M.

(vii) Let f be a function defined on J \ {a}. Then limx→a f (x) = −∞ iff

for each M < 0 there exists δ > 0 such that0 < |x − a| < δ implies f (x) < M.

I We can also mix the previous limit concepts, and define thefollowing limits.

I limx→a+ f (x) = +∞,I limx→a+ f (x) = −∞,I limx→a− f (x) = +∞,I limx→a− f (x) = −∞,

I limx→+∞ f (x) = +∞,I limx→+∞ f (x) = −∞,I limx→−∞ f (x) = +∞,I limx→−∞ f (x) = +∞.




x

y

f (x)M

−M

1M

− 1M

Figure: The function f : R \ {0} → R is given by f (x) = 1/x for all x ∈ R \ {0}. Herelimx→0+ f (x) = +∞, while limx→0− f (x) = −∞. The two-sided limit at a does not exist. Itcannot be +∞ because there is no δ such that x ∈ (−δ,+δ) implies f (x) > M.




I The next result tells us that if the left-hand and right-hand limitsat a of a function exist and are equal, then the two-sided limit at aexists and is equal to these two limits.

I It also says the converse is also true – if the limit exists, thenboth the left-hand and right-hand limits exist and are equal to thetwo-sided limit.

TheoremLet f be a function defined on J\{a} for some open interval Jcontaining a. Then limx→a f (x) exists iff the limits limx→a+ f (x) andlimx→a− f (x) both exist and are equal, in which case all three limitsare equal.




TheoremLet f1 and f2 be functions for which the limits L1 = limx→aS f1(x) andL2 = limx→aS f2(x) exist and are finite. Then

(i) limx→aS (f1 + f2)(x) = L1 + L2;(ii) limx→aS (f1f2)(x) = L1L2;(iii) limx→aS (f1/f2)(x) = L1/L2 if L2 6= 0 and f2(x) 6= 0 for x ∈ S.

Proof of (iii).blah �



Limits of Functions.I The following result provides conditions under which we can

change the order of taking the limit and applying the function g.

TheoremLet f be a function for which the limit L = limx→aS f (x) exists and isfinite. If g is a function defined on {f (x) | x ∈ S} ∪ {L} that iscontinuous at L, then limx→aS g ◦ f (x) exists and equals g(L).

ExampleCompute limx→1

√x3 + 1.

Proof.Here the function g : R+ → R, given by g(x) =

√x for all x ∈ R+ is

continuous. Thus

limx→1

√x3 + 1 =

√limx→1

x3 + 1 =√

1 + 1 =√

2. �


Documents

MTAEA Continuity and Limits of Functions · Continuity and Limits of Functions Continuous Functions. Example Let f : R !R be given by f(x) = 2x2 +1 for all x 2R. Prove f is continuous