Discussion: Givenε >0, we want to findδ >0 such that if 0<|x−2|< δ⇒ |x2−4|< ε.Now, |x2−4|=|(x−2)(x+ 2)| =|x−2||x+ 2| ≤ |x−2|(|x|+ 2)

LIMITS

DR.HAMED AL-SULAMI

Definition 0.1. LetE⊆R,and letabe a limit point ofE.For a functionf :E→R,a real numberL is said to be alimit of f ata, we write limx→af(x) =L,if, for allε >0 there existsδ >0 such that if x∈E and 0<|x−a|< δ⇒ |f(x)−L|< ε.

Note 0.1. The inequality0<|x−a|is equivalent to saying x6=a.

Example 0.1. Prove that limx→2x²= 4.

Discussion: Givenε >0, we want to findδ >0 such that if 0<|x−2|< δ⇒ |x²−4|< ε.Now,

|x²−4|=|(x−2)(x+ 2)|

=|x−2||x+ 2|

≤ |x−2|(|x|+ 2).

If we assume that|x−2|<10, then |x| −2<|x−2|<10. Hence|x| −2<10⇒ |x|<12.

Now,|x²−4| ≤ |x−2|(|x|+ 2)

≤ |x−2|(12 + 2)

≤14|x−2|.

Now, if we assume 14|x−2|< ε⇒ |x−2|< ε 14. Now, we have the following conditions on|x−2|:|x−2|<10 and|x−2|< ε

14.If we choose δ= min{10, ε 14}.

Proof. Letε >0 be given. Let δ= min{10, ε 14}.

Now, if 0<|x−2|< δ⇒ |x²−4| ≤14|x−2|

<14δ

<14.ε 14

=ε.

Thus, if 0<|x−2|< δ⇒ |x²−4|< ε.

¤

Example 0.2. Prove that limx→−1(x²+x+ 7) = 7.

Discussion:

Givenε >0,we want to findδ >0 such that if 0<|x−(−1)|=|x+ 1|< δ⇒ |x²+x+ 7−7|< ε.Now, the idea is to start with|x²+x+ 7−7|and try to make it less than or equal to (number).|x+ 1|.

|x²+x+ 7−7|=|x²+x|

=|x+ 1||x|

If we assume that|x+ 1|<1, then |x| −1<|x+ 1|<1.Hence |x| −1<1⇒ |x|<2.

Now,|x²+x+ 7−7| ≤ |x+ 1||x|

≤ |x+ 1|2

≤2|x+ 1|.

ε

Proof. Letε >0 be given. Let δ= min{1,ε 2}.

Now, if 0<|x+ 1|< δ⇒ |x²+x+ 7−7|=|x+ 1||x|

<2|x+ 1|

<2δ

<2.ε 2

=ε.

Thus, if 0<|x+ 1|< δ⇒ |x²+x+ 7−7|< ε.

¤

Example 0.3. Prove that limx→1

µ3x²−1 2x+ 1

¶

=2 3.

Discussion: Givenε >0,we want to findδ >0 such that if 0<|x−1|< δ⇒

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯< ε.Now, the idea is to start with

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯and try to make it less than or equal to (number).|x−1|.

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯=

¯¯

¯¯3(3x²−1)−2(2x+ 1) 3(2x+ 1)

¯¯

¯¯

=

¯¯

¯¯9x²−3−4x−2 3(2x+ 1)

¯¯

¯¯

=

¯¯

¯¯9x²−4x−5 3(2x+ 1)

¯¯

¯¯

= |9x²−4x−5|

3|2x+ 1|

= |9x²−9x+ 5x−5|

3|2x+ 1|

= |9x(x−1) + 5(x−1)|

3|2x+ 1|

= |(9x+ 5)(x−1)|

3|2x+ 1|

= |x−1||9x+ 5|

3|2x+ 1|

≤ |x−1|(9|x|+ 5) 3|2x+ 1|

If we assume that|x+ 1|< 1

2, then|x| −1<|x−1|< 1

2 and −1

2 < x−1<1 2. Hence |x| −1< 1

2 ⇒ |x|<3 2, and

−1

2 < x−1< 1 2 ⇒ 1

2 < x < 3 2.

Thus 2<2x+ 1⇒ 1 2x+ 1 <1

2 Now,

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯≤ |x−1|(9|x|+ 5) 3|2x+ 1|

≤ |x−1| 1

3|2x+ 1|(9|x|+ 5)

<|x−1|1 3 1 2(9(3

2) + 5).

= 37 12|x−1|

Now, if we assume 37

12|x−1|< ε⇒ |x−1|< 12ε 37. Now, we have the following conditions on|x−1|:|x−1|<1

2 and|x−1|<12ε

37.If we choose δ= min{1 2,12ε

37}.

Proof. Letε >0 be given. Let δ= min{¹₂,12ε 37}.

Now, if 0<|x+ 1|< δ⇒

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯<|x−1|1 3 1 2(9(3

2) + 5)

=37 12|x−1|

<37 12δ

<37 12.12ε

37

=ε.

Thus, if 0<|x−1|< δ⇒

¯¯

¯¯3x²−1 2x+ 1 −2

3

¯¯

¯¯< ε.

¤

Theorem 0.1. Let f :E→Rand leta be a limit point ofE. Thenlimx→af(x) =L if and only if for every sequence{xn} ⊆E such that limn→∞xn =a andxn6=a ∀n∈N, thenlimn→∞f(xn) =L.

Proof. (⇒) Suppose that limx→af(x) =L.Let{xn} ⊆Esuch that limn→∞xn=aandxn6=a∀n∈N.

We want to show that limn→∞f(xn) =L.

Letε >0 be given.

Since limx→af(x) =L, then there existδ >0 such that 0<|x−a|< δ, ⇒ |f(x)−L|< ε.

Since, limn→∞xn=a,then there existsN∈Nsuch that ifn > N ⇒ |xn−a|< δ.

Now, ifn > N ⇒ |xn−a|< δ⇒ |f(xn)−L|< ε.Hence , ifn > N ⇒ |f(xn)−L|< ε.

Thus

n→∞lim f(xn) =L.

(⇐) Suppose that for every sequence {xn} ⊆E such that limn→∞xn =a and xn 6=a ∀ n ∈N, then limn→∞f(xn) =L.Assume that limx→af(x)6=L. Then there existε0>0 such that for allδ >0 there existx∈E andx6=awith 0<|x−a|< δ,but such that|f(x)−L| ≥ε0.

For all n∈N, there existsxn ∈E and xn 6=a with 0<|xn−a|< ¹_n,but such that|f(xn)−L| ≥ε0. Hence we have a sequence{xn} ⊆ E such that limn→∞xn =a andxn 6=a ∀n∈ N, but the sequence {f(xn)}does not converges. Contradiction. Hence limx→af(x) =L. ¤

Note 0.2. The main use of this theorem is it usually used to prove the limit of some function does not exist at some limit point.

Example 0.4. Letf(x) = 1

x.Prove that limx→0 1

x does not exist.

1 2 3

-1 -2 -3

1 2 3

-1 -2 -3

Figure 1. y= 1 x

Solution:

Let{xn}={¹_n} ⊆ D(f) and limn→∞1

n = 0.Nowf(xn) =f(_n¹) = ¹1

n =n,and limn→∞f(xn) = limn→∞n=∞. Hence limx→0 1

x DNE.

Example 0.5. Letg(x) =









2, ifx≥1;

4, ifx <1

. Prove that limx→1g(x) does not exist.

1 2 3 4 5

1 2 3

-1 -2 -3

Figure 2. y=g(x) Solution:

Let{xn}={1+⁽⁻¹⁾_n ⁿ} ⊆ D(g) and limn→∞(1+⁽⁻¹⁾_nⁿ) = 1.Now,g(xn) =g³

1 +⁽⁻¹⁾_nⁿ´

=









2, ifnis even;

4, ifnis odd . Hence{g(xn)}={4,2,4,2, . . .}.Thus limn→∞g(xn) DNE. Hence limx→0g(x) DNE.

Example 0.6. Letg(x) =









x, ifx∈Q;

0, ifx∈Q^c

. Leta∈R− {0},prove that limx→ag(x) does not exist.

Solution:

Let a ∈ R− {0}. There exist two sequences {xn} ⊆ Qsuch thatlimn→∞xn = a and {yn} ⊆ Q^c such that limn→∞yn = a. Now, g(xn) = xn and g(yn) = 0. Hence limn→∞g(xn) = limn→∞xn = a and limn→∞g(yn) = limn→∞0 = 0. Since a 6= 0, then limn→∞g(yn) = 0 6= a = limn→∞g(xn). Thus limx→ag(x) DNE for alla∈R− {0}.

Definition 0.2. LetE⊆R,and letabe a limit point ofE.Letf :E→R.We say thatf is bounded on an open interval abouta, if there existsδ >0 andM >0 such that|f(x)| ≤M ∀x∈(a−δ, x+δ)∩E.

Theorem 0.2. Let f :E →Rand let abe a limit point of E. If limx→af(x) exists, thenf is bounded on some open interval abouta.

Proof. LetL= limx→af(x).Since limx→af(x) exists, then there existδ >0 such that

if x ∈ E and 0 < |x−a| < δ ⇒ |f(x)−L| < 7. Hence if x ∈ (a−δ, a+δ)∩E and x 6= a, then

|f(x)| − |L| ≤ |f(x)−L|<7.Thus ifx∈(a−δ, a+δ)∩Eandx6=a⇒ |f(x)|<7 +|L|.

Now, ifa /∈E,letM = 7 +|L|and ifa∈E letM = sup{7 +|L|,|f(a)|}.

Thus ifx∈(a−δ, a+δ)∩E ⇒ |f(x)| ≤M. ¤

Lemma 1. Iflimx→af(x) =L,then there exists δ >0 such that if 0<|x−a|< δ⇒ |L|

2 <|f(x)|<|L|+ 1.

Proof.

Since lim

x→af(x) =L,∃δ1>0 such that if 0<|x−a|< δ1⇒ |f(x)−L|<1

⇒ |f(x)| − |L| ≤ |f(x)−L|<1

⇒ |f(x)|<1 +|L|

Thus∃δ1>0 such that if 0<|x−a|< δ1⇒ |f(x)|<1 +|L|.

Also since lim

x→af(x) =L,∃δ2>0 such that if 0<|x−a|< δ2⇒ |f(x)−L|<|L|

2

⇒ |L| − |f(x)| ≤ |f(x)−L|<|L|

2

⇒ −|f(x)|<|L|

2 − |L|

⇒ −|f(x)|<−|L|

2

⇒ |L|

2 <|f(x)|

Thus∃δ2>0 such that if 0<|x−a|< δ2⇒ |L|

2 <|f(x)|.

Letδ= min{δ1, δ2}>0. Then if 0<|x−a|< δ⇒ |L|

2 <|f(x)|<1 +|L|. ¤

Theorem 0.3. Let f, g : E → R and let a be a limit point of E. Suppose that limx→af(x) = L, and limx→ag(x) =M. Then

(i) limx→a(f ±g)(x) =L±M.

(ii) limx→a(f g)(x) =LM.

(iii) If g(x)6= 0∀ x∈E andM 6= 0, thenlimx→a

µf g

¶

(x) = L M.

Proof. We will prove (i) and leave (ii) and (iii) for the homework.

Letε >0 be given. Since limx→af(x) = L,and limx→ag(x) = M, then there existδ1, δ2 >0 ifx∈E and 0 < |x−a| < δ1 ⇒ |f(x)−L| < ^ε₂ and if x ∈ E and 0 < |x−a| < δ2 ⇒ |g(x)−M| < ^ε₂. Let δ= min{δ1, δ2}>0.ifx∈E and 0<|x−a|< δ⇒ |g(x)−M|<^ε₂ and|f(x)−L|< ^ε₂.

Now, ifx∈Eand 0<|x−a|< δ⇒ |(f±g)(x)−(L±M)| ≤ |f(x)−L|+|g(x)−M|< ^ε₂+^ε₂ =ε. ¤

Theorem 0.4. Squeeze Theorem: Let f, g, h:E →Rand let abe a limit point of E.Suppose that f(x)≤h(x)≤g(x)∀ x∈E. Iflimx→af(x) =L= limx→ag(x),thenlimx→ah(x) =L.

Proof. See homework 4. ¤

Definition 0.3. LetE⊆R,and letf :E→R.

(a) Ifa∈Ris a limit point ofE∩(a,∞),then we say thatL∈Ris aright-hand limit of f ata, and we write

x→alim⁺f(x) =L,

if, for allε >0 there existsδ >0 such that if x∈E anda < x < a+δ⇒ |f(x)−L|< ε.

(b) Ifa∈Ris a limit point ofE∩(−∞, a),then we say thatL∈Ris aleft-hand limit of f ata, and we write

x→alim⁻f(x) =L,

if, for allε >0 there existsδ >0 such that if x∈E anda−δ < x < a⇒ |f(x)−L|< ε.

Example 0.7. Letg(x) =









x², ifx≥ −1;

−x²−4x, ifx <−1

. Find limx→−1⁺g(x) and limx→−1⁻g(x).

Solution:

limx→−1⁺g(x) = limx→−1⁺x²= (−1)²= 1 and

lim_x→−1−g(x) = lim_x→−1−(−x²−4x) =−(−1)²−4(−1) =−1 + 4 = 3

1 2 3

-1

1 2 3

-1 -2 -3

Figure 3. y=g(x)

Theorem 0.5. Let f : E → R and let a be a limit point of E. Then limx→af(x) = L if and only if lim_x→a⁺f(x) =L= lim_x→a⁻f(x).

Proof. (⇒) Suppose limx→af(x) =L. Let ε >0 be given. Then there exists δ >0 such that if x∈ E and 0<|x−a|< δ⇒ |f(x)−L|< ε.

Now, if 0<|x−a|< δ ⇒a−δ < x < a+δ.

Ifa < x < a+δ⇒0<|x−a|< δ ⇒ |f(x)−L|< ε.Hence limx→a⁺f(x) =L.

Ifa−δ < x < a⇒0<|x−a|< δ ⇒ |f(x)−L|< ε.Hence limx→a⁻f(x) =L.

(⇐) Suppose lim_x→a+f(x) = L= lim_x→a−f(x). Let ε >0 be given. Then there exist δ1, δ2 >0 such that if x ∈ E and a−δ1 < x < a ⇒ |f(x)−L| < ε and a < x < a+δ2 ⇒ |f(x)−L| < ε. Let δ= min{δ1, δ2}.Now, if 0<|x−a|< δ⇒a−δ1< a−δ < x < a+δ < a+δ2⇒ |f(x)−L|< ε.Hence

limx→af(x) =L. ¤