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∈Rsuch 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 M1+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+1/2for somen∈Nthen f(x) =1, while ifx=2n+3/2then 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=1or 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→Ris 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{xn}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 limxn=∞. 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→Rbe a function and supposeShas∞as a cluster point. We say f(x) diverges to infinityasxgoes to∞, if for everyN∈Rthere exists anM∈Rsuch 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 iflimx→∞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 xnis 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