Continuous Functions
3.1.4 Limits of restrictions and one-sided limits Sometimes we work with the function defined on a subset.
Definition 3.1.14. Let f: S→Rbe a function. LetA⊂S. Define the function f|A: A→Rby f|A(x):= f(x) forx∈A.
The function f|A is called therestrictionof f toA.
The function f|A is simply the function f taken on a smaller domain. The following proposition is the analogue of taking a tail of a sequence.
Proposition 3.1.15. Let S⊂R, c∈R, and let f: S→Rbe a function. Suppose A⊂S is such that there is someα >0such that(A\ {c})∩(c−α,c+α) = (S\ {c})∩(c−α,c+α).
(i) The point c is a cluster point of A if and only if c is a cluster point of S.
(ii) Supposing c is a cluster point of S, then f(x)→L as x→c if and only if f|A(x)→L as x→c. Proof. First, letcbe a cluster point ofA. SinceA⊂S, then if(A\ {c})∩(c−ε,c+ε)is nonempty
for everyε >0, then(S\ {c})∩(c−ε,c+ε) is nonempty for every ε >0. Thus cis a cluster
point ofS. Second, supposecis a cluster point ofS. Then forε >0 such thatε <α we get that
(A\ {c})∩(c−ε,c+ε) = (S\ {c})∩(c−ε,c+ε), which is nonempty. This is true for allε<α
and hence(A\ {c})∩(c−ε,c+ε)must be nonempty for allε>0. Thuscis a cluster point ofA.
Now suppose f(x)→Lasx→c. That is, for everyε>0 there is aδ >0 such that ifx∈S\ {c}
and|x−c|<δ, then|f(x)−L|<ε. BecauseA⊂S, ifxis inA\ {c}, thenxis inS\ {c}, and hence
f|A(x)→Lasx→c.
Finally suppose f|A(x)→Lasx→c. For everyε>0 there is aδ0>0 such that ifx∈A\{c}and
|x−c|<δ0, thenf|A(x)−L
<ε. Takeδ :=min{δ0,α}. Now supposex∈S\{c}and|x−c|<δ.
As|x−c|<α, thenx∈A\ {c}, and as|x−c|<δ0, we have|f(x)−L|=f|A(x)−L <ε.
The hypothesis of the proposition is necessary. For an arbitrary restriction we generally only get implication in only one direction, see .
The usual notation for the limit is lim x→c x∈A
f(x):=lim
x→cf|A(x).
The most common use of restriction with respect to limits are theone-sided limits .
Definition 3.1.16. Let f: S→Rbe function and letcbe a cluster point ofS∩(c,∞). Then if the limit of the restriction of f toS∩(c,∞)asx→cexists, define
lim
x→c+f(x):=xlim→cf|S∩(c,∞)(x).
Similarly, ifcis a cluster point ofS∩(−∞,c)and the limit of the restriction asx→cexists, define lim
x→c−f(x):=limx→cf|S∩(−∞,c)(x).
∗There are a plethora of notations for one sided limits. E.g. for lim
x→c−f(x)one sees limx→c x<c
f(x), lim
The proposition above does not apply to one-sided limits. It is possible to have one-sided limits, but no limit at a point. For example, define f: R→Rby f(x):=1 forx<0 and f(x):=0 for
x≥0. We leave it to the reader to verify that lim
x→0−f(x) =1, xlim→0+f(x) =0, xlim→0f(x) does not exist.
We have the following replacement.
Proposition 3.1.17. Let S⊂R be a set such that c is a cluster point of both S∩(−∞,c) and
S∩(c,∞), and let f: S→Rbe a function. Then c is a cluster point of S and
lim
x→cf(x) =L if and only if xlim→c−f(x) =xlim→c+ f(x) =L.
That is, a limit exists if both one-sided limits exist and are equal, and vice-versa. The proof is a straightforward application of the definition of limit and is left as an exercise. The key point is that
S∩(−∞,c)
∪ S∩(c,∞)
=S\ {c}.
3.1.5
Exercises
Exercise3.1.1: Find the limit or prove that the limit does not exist
a)lim x→c √ x, for c≥0 b)lim x→cx 2+x+1, for any c∈ R c)lim x→0x 2cos(1/x) d)lim
x→0sin(1/x)cos(1/x) e)xlim→0sin(x)cos(1/x) Exercise3.1.2: Prove .
Exercise3.1.3: Prove . Exercise3.1.4: Prove .
Exercise3.1.5: Let A⊂S. Show that if c is a cluster point of A, then c is a cluster point of S. Note the
difference from .
Exercise3.1.6: Let A⊂S. Suppose c is a cluster point of A and it is also a cluster point of S. Let f:S→R
be a function. Show that if f(x)→L as x→ c, then f|A(x)→ L as x→c. Note the difference from
.
Exercise 3.1.7: Find an example of a function f: [−1,1]→R such that for A:= [0,1], the restriction
f|A(x)→0as x→0, but the limit of f(x)as x→0does not exist. Note why you cannot apply
.
Exercise3.1.8: Find example functions f and g such that the limit of neither f(x)nor g(x)exists as x→0, but such that the limit of f(x) +g(x)exists as x→0.
Exercise3.1.9: Let c1be a cluster point of A⊂Rand c2 be a cluster point of B⊂R. Suppose f:A→B
and g:B→Rare functions such that f(x)→c2as x→c1and g(y)→L as y→c2. If c2∈B also suppose that g(c2) =L. Let h(x):=g f(x)and show h(x)→L as x→c1. Hint: Note that f(x)could equal c2for many x∈A, see also .
Exercise3.1.10: Let c be a cluster point of A⊂R, and f:A→Rbe a function. Suppose for every sequence
{xn}in A, such thatlimxn=c, the sequence{f(xn)}∞n=1is Cauchy. Prove thatlimx→c f(x)exists.
Exercise3.1.11: Prove the following stronger version of one direction of : Let S⊂R, c be a
cluster point of S, and f: S→Rbe a function. Suppose that for every sequence{xn}in S\ {c}such that
limxn=c the sequence{f(xn)}is convergent. Then show f(x)→L as x→c for some L∈R. Exercise3.1.12: Prove .
Exercise3.1.13: Suppose S⊂Rand c is a cluster point of S. Suppose f:S→Ris bounded. Show that
there exists a sequence{xn}with xn∈S\ {c}andlimxn=c such that{f(xn)}converges.
Exercise3.1.14(Challenging): Show that the hypothesis that g(c2) =L in is necessary. That is, find f and g such that f(x)→c2 as x→c1 and g(y)→L as y→c2, but g f(x)does not go to L as x→c1.
Exercise3.1.15: Show that the condition of being a cluster point is necessary to have a reasonable definition
of a limit. That is, suppose c is not a cluster point of S⊂R, and f:S→Ris a function. Show that every L
would satisfy the definition of limit at c without the condition on c being a cluster point.
Exercise3.1.16:
a) Prove .
