** Continuous Functions**

**3.5 Limits at infinity**

Note: less than 1 lecture (optional, can safely be omitted unless or is also covered)

### 3.5.1

### Limits at infinity

As for sequences, a continuous variable can also approach infinity. Let us make this notion precise.
Definition 3.5.1. We say∞is a cluster point of S_{⊂}_{R}, if for every M_{∈}_{R}, there exists anx_{∈}S
such thatx≥M. Similarly,−∞is a cluster point ofS⊂R, if for everyM∈R, there exists anx∈S

such thatx_{≤}M.

Let f: S_{→}Rbe a function, where∞is a cluster point ofS. If there exists anL∈Rsuch that

for everyε>0, there is anM∈_{R}such that

|f(x)_{−}L|<ε

wheneverx_{∈}Sandx_{≥}M, then we say f(x)convergestoLasxgoes to∞. We callLthelimitand
write

lim

x→∞f(x):=L.
Alternatively we write f(x)_{→}Lasx_{→}∞.

Similarly, if−∞is a cluster point ofSand there exists anL∈Rsuch that for everyε>0, there

is anM∈Rsuch that

|f(x)−L|<ε

wheneverx_{∈}Sandx_{≤}M, then we say f(x)convergestoLasxgoes to−∞. We callLthelimit
and write

lim

x→−∞f(x):=L. Alternatively we write f(x)→Lasx→ −∞.

We cheated a little bit again and saidthelimit. We leave it as an exercise for the reader to prove the following proposition.

Proposition 3.5.2. The limit at∞or_{−}∞as defined above is unique if it exists.
Example 3.5.3: Let f(x):=_{|}_{x}_{|}1_{+1}. Then

lim

x→∞f(x) =0 and x→−lim∞f(x) =0.

Proof: Let ε >0 be given. Find M > 0 large enough so that _{M}1_{+1} <ε. If x≥ M, then

1

x+1 ≤ M1+1 <ε. Since _{|}_{x}_{|}1_{+1} >0 for allxthe first limit is proved. The proof for−∞is left to the

Example 3.5.4: Let f(x):=sin(πx). Then limx→∞f(x)does not exist. To prove this fact note that
ifx=2n+1_{/}_{2}_{for some}_{n}_{∈}_{N}_{then} _{f}_{(x) =}_{1, while if}_{x}_{=}_{2}_{n}_{+}3_{/}_{2}_{then} _{f}_{(x) =}_{−}_{1, so they cannot}
both be within a smallε of a single real number.

We must be careful not to confuse continuous limits with limits of sequences. We could say lim

n→∞sin(πn) =0, but xlim→∞sin(πx)does not exist.

Of course the notation is ambiguous: Are we thinking of the sequence{sin(πn)}∞_{n}_{=1}or the function

sin(πx)of a real variable? We are simply using the convention thatn∈_{N}, whilex∈_{R}. When the

notation is not clear, it is good to explicitly mention where the variable lives, or what kind of limit are you using. If there is possibility of confusion, one can write, for example,

lim n→∞

n∈N

sin(πn).

There is a connection of continuous limits to limits of sequences, but we must take all sequences going to infinity, just as before in .

Lemma 3.5.5. Suppose f: S_{→}_{R}is a function,∞is a cluster point of S_{⊂}_{R}, and L_{∈}_{R}. Then
lim

x→∞f(x) =L if and only if

lim

n→∞f(xn) =L
for all sequences_{{}x_{n}}in S such that lim

n→∞xn=∞.

The lemma holds for the limit asx→ −∞. Its proof is almost identical and is left as an exercise.
Proof. First suppose f(x)_{→}Lasx_{→}∞. Given anε >0, there exists anMsuch that for allx≥M

we have|f(x)−L|<ε. Let{xn}be a sequence inSsuch that limx_{n}=∞. Then there exists anN

such that for alln≥N we havexn≥M. And thus|f(xn)−L|<ε.

We prove the converse by contrapositive. Suppose f(x)does not go toLasx→∞. This means
that there exists anε >0, such that for everyn∈_{N}, there exists anx∈S,x≥n, let us call itxn,
such that|f(xn)−L| ≥ε. Consider the sequence{xn}. Clearly{f(xn)}does not converge toL. It
remains to note that limxn=∞, becausexn≥nfor alln.

Using the lemma, we again translate results about sequential limits into results about continuous limits asxgoes to infinity. That is, we have almost immediate analogues of the corollaries in . We simply allow the cluster pointcto be either∞or−∞, in addition to a real number. We leave it to the student to verify these statements.

### 3.5.2

### Infinite limit

Just as for sequences, it is often convenient to distinguish certain divergent sequences, and talk about limits being infinite almost as if the limits existed.

Definition 3.5.6. Let f: S_{→}_{R}be a function and supposeShas∞as a cluster point. We say f(x)
diverges to infinityasxgoes to∞, if for everyN_{∈}_{R}there exists anM_{∈}_{R}such that

f(x)>N

wheneverx_{∈}Sandx_{≥}M. We write

lim

x→∞f(x):=∞,
or we say that f(x)_{→}∞asx_{→}∞.

A similar definition can be made for limits asx_{→ −}∞or asx_{→}cfor a finitec. Also similar
definitions can be made for limits being−∞. Stating these definitions is left as an exercise. Note
that sometimesconverges to infinityis used. We can again use sequential limits, and an analogue of

is left as an exercise.

Example 3.5.7: Let us show that limx→∞1+x

2

1+x =∞.
Proof: Forx_{≥}1 we have

1+x2 1+x ≥ x2 x+x = x 2.

GivenN_{∈}R, takeM=max{2N+1,1}. Ifx≥M, thenx≥1 andx/2>N. So

1+x2 1+x ≥

x 2 >N.

### 3.5.3

### Compositions

Finally, just as for limits at finite numbers we can compose functions easily.

Proposition 3.5.8. Suppose f: A_{→}B, g: B_{→}_{R}, A,B_{⊂}R, a∈R∪ {−∞,∞}is a cluster point of

A, and b∈R∪ {−∞,∞}is a cluster point of B. Suppose

lim

x_{→}af(x) =b and ylim→bg(y) =c
for some c∈R∪ {−∞,∞}. If b∈B, then suppose g(b) =c. Then

lim

x→ag f(x)

=c.

The proof is straightforward, and left as an exercise. We already know the proposition when
a,b,c_{∈}R, see Exercises and . Again the requirement thatgis continuous atb, ifb∈B,

is necessary.

Example 3.5.9: Leth(x):=e−x2+x. Then lim

x→∞h(x) =0. Proof: The claim follows once we know

lim
x→∞−x
2_{+}_{x}_{=}_{−}_{∞}
and
lim
y→−∞e
y_{=}_{0}_{,}

### 3.5.4

### Exercises

Exercise3.5.1: Prove .

Exercise 3.5.2: Let f: [1,∞)→R be a function. Define g: (0,1]→R via g(x):= f(1/x). Using the

definitions of limits directly, show thatlimx→0+g(x)exists if and only iflim_{x}_{→}_{∞}f(x)exists, in which case

they are equal.

Exercise3.5.3: Prove .

Exercise3.5.4: Let us justify terminology. Let f:R→Rbe a function such thatlimx→∞f(x) =∞(diverges

to infinity). Show that f(x)diverges (i.e. does not converge) as x_{→}∞.

Exercise 3.5.5: Come up with the definitions for limits of f(x) going to−∞as x→∞, x→ −∞, and as

x→c for a finite c∈R. Then state the definitions for limits of f(x)going to∞as x→ −∞, and as x→c for

a finite c∈R.

Exercise 3.5.6: Suppose P(x):=xn+an−1xn−1+···+a1x+a0 is a monic polynomialof degree n≥1
(monic means that the coefficient of xn_{is 1).}

a) Show that if n is even thenlimx→∞P(x) =limx→−∞P(x) =∞.

b) Show that if n is odd thenlimx→∞P(x) =∞andlimx→−∞P(x) =−∞(see previous exercise).

Exercise3.5.7: Let_{{}xn}be a sequence. Consider S:=N⊂R, and f:S→Rdefined by f(n):=xn. Show

that the two notions of limit,

lim

n→∞xn and xlim→∞f(x)

are equivalent. That is, show that if one exists so does the other one, and in this case they are equal.

Exercise3.5.8: Extend as follows. Suppose S⊂Rhas a cluster point c∈R, c=∞, or c=−∞.

Let f:S→Rbe a function and let L=∞or L=−∞. Show that

lim

x→cf(x) =L if and only if nlim→∞f(xn) =L for all sequences{xn}such thatlimxn=c. Exercise 3.5.9: Suppose f: R→ R is a 2-periodic function, that is f(x+2) = f(x) for all x. Define

g:R→Rby
g(x):= f
√
x2_{+}_{1}_{−}_{1}
x
!

a) Find the functionϕ:(−1,1)→Rsuch that g ϕ(t)= f(t), that isϕ−1(x) =

√

x2_{+}_{1}_{−}_{1}

x .

b) Show that f is continuous if and only if g is continuous and lim