Download Definition: lim f(x) = L means: (1) f is defined on an open interval

Definition:
lim f (x) = L means:
x→a
(1) f is defined on an open interval containing a but not necessarily at a, and
(2) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in
the domain of f , if 0 < |x − a| < δ, then |f (x) − L| < ǫ.
In this case we say that the limit of f at a is L.
Here is a graphical explanation of the definition. First of all, you have to make a
clever guess at what the limit L might be. Mark this L on the y-axis. Next, we are given
a positive real number ǫ – over which we have no control as we need to prove a statement
for all positive ǫ. Draw a horizontal swath around y = L going ǫ units up and ǫ units
down. In the picture below this swath is in light gray.
L+ǫ
L
L−ǫ
a
Now the burden is on you to show that there exists a vertical swath around the vertical
line x = a going δ units to the left and δ units to the right such that for any x in that
swath that is not equal to a, f (x) falls into the horizontal ǫ swath.
L+ǫ
L
L−ǫ
a−δ a a+δ
The thickened part of the graph now shows clearly that for all x within δ of a, f (x)
falls in the given ǫ-swath.
Clearly if ǫ gets smaller, δ has to get smaller too. But if ǫ gets larger, we may keep
the old δ.
In any case, the order of words in a definition matters. First of all, f , a, and
L are given ahead of time. Then, the first-named quantity, ǫ, is chosen/given first, then δ
depends on ǫ, f and a, after which xs are arbitrary depending on δ and a.
Most of the time we will be giving algebraic proofs of limits, but it can help your proof
if you keep in mind a geometric picture.
1
Asking for the limit of f at a is not the same thing as asking for f (a). The following
three functions all have the same limit L at a; in the leftmost case, the limit equals f (a),
in the middle case, the function is not defined at a, and in the rightmost case, the function
is defined at a but f (a) 6= L.
L
L
L
a
a
a
Handy facts and tricks:
•
•
a
ba
b
c
d
+ dc =
= ad
bc .
ad+bd
bd .
√ √
√
√
• (a − b)(a + b) = a2 − b2 . In particular, ( a − b)( a + b) = a − b.
• (a + b)2 = a2 + 2ab + b2 .
• (a + b)3 = a3 + 3a2 b + 3ab2 + b3 .
• |ab| = |a| · |b|.
• If a ≤ b then a + c ≤ b + c. If a < b then a + c < b + c.
• If a ≤ b and c ≥ 0, then ac ≤ bc.
• If a < b and c > 0, then ac < bc.
• If a ≤ b and c < 0, then ac ≥ bc.
• If a < b and c < 0, then ac > bc.
• Triangle inequality: For all real numbers a and b, |a ± b| ≤ |a| + |b|.
• In general, if |a| < c you cannot conclude that |a + b| < c + b. (Counterexample:
a = 2, b = −5, c = 3.)
• Reverse triangle inequality: For all real numbers a and b, |a ± b| ≥ |a| − |b|, and
similarly, |a ± b| ≥ |b| − |a|.
• Add a clever 0.
• Multiply by a clever 1.
• Informal goal #1 for proving epsilon-delta limits: Find “something” with the property
that |f (x) − L| ≤ something · |x − a|.
Informal goal #2: Find a positive real number B such that “something” from goal #1
is less than or equal to B.
Then make sure that δ ≤ ǫ/B.
2
How to prove things:
Statement
How to prove it
For all x of some type, property
P holds for x.
There exists x of some type such
that property P holds for x.
Let x be arbitrary of the specified type. Prove
that property P holds for x.
Find/construct/determine a concrete x of the
specified type. Prove that property P holds
for x.
Suppose that P is true. Then prove Q.
If P then Q.
How does this proving scheme apply to limits?
Let P be the property [of ǫ] that there exists a real number δ > 0 such that for all
x in the domain of f , if 0 < |x − a| < δ, then |f (x) − L| < ǫ. This is quite a mouthful
of a property P ! For proving mouthfuls, it is easier to break them into manageble pieces,
and by giving them an abbreviated name, such as “P ” rather than the mouthful, it may
be easier to think about it.
In order to prove that every ǫ > 0 has property P , as in the chart above, the proof
must start with “Let ǫ be an arbitrary positive real number.” This is somewhat verbose,
and we will abbreviate it (standardly) to “Let ǫ > 0.” We did not change the meaning with
this abbreviation. After this start, according to the proving chart above, we have to prove
that P holds for this (arbitrary) ǫ. In other words, we have to prove that there exists a
positive real δ such that et cetera. By the chart, we have to construct/find a δ first, and
then prove the specifying property for δ, i.e., that for all x in the domain, et new cetera.
Well, slow down, how does one construct/find a δ with the specifying property? Scratch
paper comes in very handy for this, and do make use of scratch paper! But really, the best
way to learn how to construct/find δ is through examples.
3
Examples:
In examples below, the normal font shows a possible expected write-up, and the allcaps font is showing the thoughts that should be going on in your head at the same time
and that should not be recorded in your write-up.
Example 1 lim (4x − 5) = 7.
x→3
Proof. The function x 7→ 4x − 5 is a polynomial function, so defined for all real numbers.
[(This satisfies the first condition of limits.)] Let ǫ > 0. [(We broke up the
part: For all ǫ > 0 something-or-other holds.)] Set δ = ǫ/4. [(The somethingor-other above said that there exists δ > 0 such that somethingelse holds.
So, we construct an explicit δ (my experience tells me that ǫ/4 works, but
you may not have such experience quite yet; read the proof in Example 2
where YOU will have to figure that out.)] Then δ is a positive real number.
[(Now we have to prove that for all x, if 0 < |x − 3| < δ, then |f (x) − L| < ǫ.
So: “Let x be an arbitrary real number. Assume that 0 < |x − 3| < δ.” is
written in one shorter sentence next:)] Let x be any real number that satisfies
0 < |x − 3| < δ. Then
|(4x − 5) − 7| = |4x − 12|
= 4|x − 3|
< 4δ
ǫ
=4
4
= ǫ.
Example 2
lim (4x2 − 5x + 2) = 28.
x→−2
Proof. The function x 7→ 4x2 − 5x + 2 is a polynomial function, so defined for all real
[(to be determined still; your final
numbers. Let ǫ > 0. Set δ =
write-up will have it all filled in, but as we are writing the proof, we don’t
know δ yet!)]. Then δ is a positive real number. Let x be any real number that satisfies
0 < |x + 2| < δ. Then
|(4x2 − 5x + 2) − 28| = |4x2 − 5x − 26| [(x − a better be a factor of this!)]
= |(x + 2)(4x − 13)|
= |4x − 13| · |x + 2| [(Goal #1 accomplished: sth times |x − a|)]
4
[Goal #2: we want to bound sth above by a constant.
But obviously |4x − 13| is not bounded above by a constant for all x. So we need to make a restriction on
δ: let’s make sure that δ is at most 1 (or 2, or 15, it
doesn’t matter what positive number you pick in this
example).] [So on the δ line write: δ = min{1,
= |4(x + 2 − 2) − 13| · |x + 2| [(adding a clever zero)]
}.]
= |4(x + 2) − 21| · |x + 2|
≤ (|4(x + 2)| + 21) · |x + 2| (by triangle inequality |a ± b| ≤ |a| + |b|)
< (4 · 1 + 21) · |x + 2| (since δ ≤ 1 and |x + 2| < δ)
= 25 · |x + 2| [(Goal #2 accomplished: constant times |x − a|)]
[Goal #2: Now go back to δ and finish with: δ =
min{1, ǫ/constant}.]
< 25 · δ
≤ 25 · ǫ/25
= ǫ.
Example 3
lim (4/x2 ) = 4.
x→−1
Proof. The function x 7→ 4/x2 is defined for all non-zero real numbers. Let ǫ > 0. Set
δ =
. Then δ is a positive real number. Let x be any real number that
satisfies 0 < |x + 1| < δ. Then
4 − 4x2 2
|4/x − 4| = x2 1 − x
= 4 2 |1 + x|[(Goal #1 accomplished: sth times |x − a|)]
x
[On the δ line write: δ = min{0.4,
}.] [Check that
1 in place of 0.4 would NOT work here. Any number
strictly between 0 and 1 works.]
2 − (x + 1) |1 + x|
= 4
x2
2 + |x + 1|
≤4
|1 + x| (by triangle inequality)
x2
2.4
≤ 4 2 |1 + x| (since δ ≤ 0.4)
x
5
9.6
|1 + x|
x2
9.6
|1 + x| (because x is within 0.4 of −1, hence at least 0.6 away from 0)
≤
0.62
[On the δ line now write: δ = min{0.4, 0.62ǫ/9.6}.]
9.6
<
δ
0.62
9.6
0.62 ǫ/9.6
≤
2
0.6
= ǫ.
=
Example 4 lim (x2 − 2x) = a2 − 2a.
x→a
Proof. Any a is a limit point of the domain of the given polynomial function. Let ǫ > 0.
Set δ = min{1, ǫ/(1 + |2a − 2|)}. Let x satisfy 0 < |x − a| < δ. Then
|(x2 − 2x) − (a2 − 2a)| = |(x2 − a2 ) − (2x − 2a)| (by algebra)
= |(x + a)(x − a) − 2(x − a)| (by algebra)
= |(x + a − 2)(x − a)| (by algebra)
= |x + a − 2| |x − a|
= |(x − a) + 2a − 2| |x − a| (by adding a clever 0)
≤ (|x − a| + |2a − 2|) |x − a| (by triangle inequality)
≤ (1 + |2a − 2|) |x − a| (since |x − a| < δ ≤ 1)
< (1 + |2a − 2|)δ
≤ (1 + |2a − 2|)ǫ/(1 + |2a − 2|)
= ǫ.
6
Exercises to go with epsilon-delta proofs and Section 1.8:
1 Fill in the blanks of the following proof that limx→2 (x2 − 3x) = −2. Explain why none
of the inequalities can be changed into equalities.
Let ǫ > 0. Set δ =
. Let x satisfy 0 < |x − 2| < δ. Then
|(x2 − 3x) − (−2)| = x2 − 3x + 2
= |x − 1| |x − 2|
(because
< (3 + 1) |x − 2|
(because
≤ (|x − 2| + 1) |x − 2|
< 4δ
ǫ
≤4
4
= ǫ.
)
)
(because
)
(because
)
(because
)
2 Determine the following limits and prove them with epsilon-delta proofs.
i) lim (x3 − 4).
x→1
1
.
x→2 x
lim xx−4
2 +2 .
x→3 √
ii) lim
iii)
iv) lim
x→4
x + 5.
x2 −9
.
x→3 x−3
lim x−3
.
2
x→3 x −9
v) lim
vi)
3 Let b be a real number and let f, g be functions defined for all real numbers a follows:
f (x) =
x3 − 4x2 ,
b,
if x 6= 5;
if x = 5,
g(x) =
x3 − 4x2 ,
b,
if x = 5;
if x =
6 5.
Prove that the limit of f (x) as x approaches 5 is independent of b, but that the limit of
g(x) as x approaches 5 depends on b.
4 Prove that limx→a (mx + l) = ma + l, where m and l are constants.
When a limit is not L
Suppose that f is defined on an open interval containing a but not necessarily at a. Then
lim f (x) 6= L means:
x→a
NOT For all real numbers ǫ > 0 there exists a real number δ > 0 such
that for
all x in the domain of f , if 0 < |x − a| < δ then |f (x) − L| < ǫ.
= There exists a real number ǫ > 0 such that NOT there exists a real number
δ > 0 such that
for all x in the domain of f , if 0 < |x − a| < δ then
|f (x) − L| < ǫ.
= There exists a real number ǫ > 0 such that for all real numbers δ > 0,
NOT for all x in the domain of f , if 0 < |x − a| < δ then |f (x) − L| < ǫ.
= There exists a real number ǫ > 0 such that for all real numbers δ > 0, there
exists x in the domain of f such that NOT if 0 < |x − a| < δ then
|f (x) − L| < ǫ.
= There exists a real number ǫ > 0 such that for all real numbers δ > 0, there
exists x in the domain of f such that 0 < |x − a| < δ and
NOT |f (x) − L| < ǫ.
= There exists a real number ǫ > 0 such that for all real numbers δ > 0, there
exists x in the domain of f such that 0 < |x − a| < δ and |f (x) − L| ≥ ǫ.
x
Example 1 The limit of |x|
as x approaches 0 does not exist. In other words, for all real
x
numbers L, limx→0 |x| 6= L.
x
Proof. The domain of the function that takes x to |x|
is the set of all non-zero real numbers,
so that 0 is a limit point of the domain. Set ǫ = 1 (Hint: half of the gap). Let δ > 0 be
an arbitrary positive number. Let x = −δ/2 if L ≥ 0, and let x = δ/2 otherwise. In both
cases, 0 < |x| = |x − 0| = δ/2 < δ. If L ≥ 0, then
−δ/2
x
|x| − L = | − δ/2| − L = | − 1 − L| = 1 + L ≥ 1 = ǫ,
and if L < 0, then
x
δ/2
=
= |1 − L| = 1 − L > 1 = ǫ.
−
L
−
L
|x|
|δ/2|
This proves the claim of the example.
Example 2 For f be given by the graph below, lim f (x) does not exist because of the
x→2
jump in the function at 2.
2
1
1
2
3
Here is an epsilon-delta proof. Say that the limit exists. Call it L.
Set ǫ = 41 . Let δ
be an arbitrary positive number. If L ≥ 32 , set x = 2 + min 14 , 2δ , and if L < 32 , set
x = 2 − min{ 14 , 2δ }. In either case,
0 < |x − 2| = min
1 δ
,
4 2
≤
δ
< δ.
2
If L ≥ 23 , by our choice x = 2 + min{ 41 , δ2 }, so that f (x) = 1 + min{ 41 , δ2 } ≤ 1 + 41 ,
whence |f (x) − L| ≥ 41 = ǫ. Similarly, if L < 23 , by our choice x = 2 − min{ 41 , δ2 }, so that
f (x) = 2 − min{ 41 , 2δ } ≥ 2 − 14 = 1 + 34 , whence |f (x) − L| ≥ 41 = ǫ. Thus no L works, so
the limit of f (x) as x approaches 2 does not exist.
Exercise 3 Below you will find an attempt at a “proof” that limx→3 x2 = 9. Explain how
the two starred steps below contain three mistakes.
Let ǫ > 0. Set δ =∗ ǫ/|x + 3|. Then
|x2 − 9| = |x − 3| · |x + 3|
<∗ δ|x + 3|
= ǫ.
Definition: The right-sided limit of f at a is L: lim f (x) = L means:
x→a+
(1) f is defined on an open interval of the form (a, b) for some b > a, and
(2) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in
the domain of f , if 0 < x − a < δ, then |f (x) − L| < ǫ.
Definition: The left-sided limit of f at a is L: lim f (x) = L means:
x→a−
(3) f is defined on an open interval of the form (a, b) for some b > a, and
(4) for every real number ǫ > 0 there exists real number δ > 0 such that for all x in
the domain of f , if 0 < a − x < δ, then |f (x) − L| < ǫ.
Example 4 lim+
x→3
√
2x − 6 = 0.
Proof. The domain here is all x ≥ 3. So 3 is a limit point of the domain. Let ǫ > 0. Set
δ = ǫ2 /2. Let x satisfy 0 < x − 3 < δ. Then
√
2x − 6 =
<
=
√
√
√
2·
2·
√
√
x−3
δ
p
2 · ǫ2 /2
= ǫ.
x2 + 4,
Example 5 Let f be given by f (x) =
x − 2,
lim f (x) = −3.
if x > 1;
if x ≤ 1.
Then lim+ f (x) = 5,
x→1−
Proof. Let ǫ > 0.
Set δ = min{1, ǫ/3}. Let x satisfy 0 < x − 1 < δ. Then
f (x) − 5 = x2 + 4 − 5 (since x > 1)
= x2 − 1
= (x + 1)(x − 1)
< (x + 1)δ (since x > 1, so x + 1 is positive)
= (x − 1 + 2)δ (by rewriting)
< (1 + 2)δ (since x − 1 < δ ≤ 1)
≤ 3ǫ/3
= ǫ.
This proves that lim f (x) = 5.
x→1+
x→1
Set δ = ǫ. Let x satisfy 0 < 1 − x < δ. Then
f (x) − (−3) = x − 2 + 3 (since x < 1)
=x−1
<δ
= ǫ.
This proves that lim f (x) = −3.
x→1−
Exercises to go with one-sided limits:
1 Let f be given by f (x) =
x
|x| .
Prove that limx→0+ f (x) = 1 and that limx→0− f (x) = −1.
2 Find a function f defined for all real numbers such that limx→0− f (x) = 2 and such that
limx→0+ f (x) = −5.
3 Find a function f defined for all real numbers such that limx→0− f (x) = 2,
limx→0+ f (x) = −5, limx→1− f (x) = 3, and limx→1+ f (x) = 0. (Try to define such
a function with fewest possible words or symbols, but do use full grammatical sentences.)