MATH1010 University Mathematics - math.cuhk.edu.hk · PDF filefrom the set of positive...
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Limits
MATH1010 University Mathematics
Department of MathematicsThe Chinese University of Hong Kong
MATH1010 University Mathematics
Limits
1 LimitsSequences
Limits of sequencesSqueeze theoremMonotone convergence theorem
Limits and Continuity
Limits of functionsExponential, logarithmic and trigonometric functionsContinuity of functions
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Infinite sequence of real numbers)
An infinite sequence of real numbers is defined by a functionfrom the set of positive integers Z+ = {1, 2, 3, . . . } to the set ofreal numbers R.
Example (Sequences)
Arithmetic sequence: an = 3n+ 4; 7, 10, 13, 16 . . .
Geometric sequence: an = 3 · 2n; 6, 12, 24, 48 . . .
Fibonacci’s sequence:
Fn =1√5
((1 +√5
2
)n
−
(1−√5
2
)n);
1, 1, 2, 3, 5, 8, 13, . . .
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Limit of sequence)
1 Suppose there exists real number L such that for any ε > 0, thereexists N ∈ N such that for any n > N , we have |an − L| < ε. Thenwe say that an is convergent, or an converges to L, and write
limn→∞
an = L.
Otherwise we say that an is divergent.
2 Suppose for any M > 0, there exists N ∈ N such that for anyn > N , we have an > M . Then we say that an tends to +∞ as ntends to infinity, and write
limn→∞
an = +∞.
We define an tends to −∞ in a similar way. Note that an isdivergent if it tends to ±∞.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example (Intuitive meaning of limits of infinite sequences)
an First few terms Limit
1
n21,
1
4,1
9,1
16, . . . 0
n
n+ 1
1
2,2
3,3
4,4
5, . . . 1
(−1)n+1 1,−1, 1,−1, . . . does not exist
2n 2, 4, 6, 8, . . . does not exist/+∞(1 +
1
n
)n
2,9
4,64
27,625
256, . . . e ≈ 2.71828
Fn+1
Fn1, 2,
3
2,5
3, . . .
1 +√5
2≈ 1.61803
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Monotonic sequence)
1 We say that an is monotonic increasing (decreasing) if forany m < n, we have am ≤ an (am ≥ an). We say that an ismonotonic if an is either monotonic increasing or monotonicdecreasing.
2 We say that an is strictly increasing (decreasing) if for anym < n, we have am < an (am > an).
Definition (Bounded sequence)
We say that an is bounded if there exists real number M suchthat |an| < M for any n ∈ N.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example (Bounded and monotonic sequence)
an Terms Bounded MonotonicConvergent
(Limit)
1
n21,
1
4,1
9,1
16, . . . X X X (0)
1− (−1)n
n2,
1
2,4
3,3
4, . . . X × X (1)
n2 1, 4, 9, 16, . . . × X ×1− (−1)n 2, 0, 2, 0, . . . X × ×(−1)nn −1, 2,−3, 4, . . . × × ×
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
If an is convergent, then an is bounded.
Convergent⇒ Bounded
Note that the converse of the above statement is not correct.
Bounded 6⇒ Convergent
The following theorem is very important and we will discuss it indetails later.
Theorem (Monotone convergence theorem)
If an is bounded and monotonic, then an is convergent.
Bounded and Monotonic⇒ Convergent
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose limn→∞
an = a and limn→∞
bn = b. Then
limn→∞
(an ± bn) = a± b.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose limn→∞
an = a and c is a real number. Then
limn→∞
can = ca.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = a and limn→∞
bn = b, then
limn→∞
anbn = ab.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = a and limn→∞
bn = b, then
limn→∞
anbn
=a
b.
Answer: F
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = a and limn→∞
bn = b 6= 0, then
limn→∞
anbn
=a
b.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = 0, then
limn→∞
anbn = 0.
Answer: F
Example
For an =1
nand bn = n, we have lim
n→∞an = 0 but
limn→∞
anbn = limn→∞
1
n· n = lim
n→∞1 = 1 6= 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = 0 and bn is convergent, then
limn→∞
anbn = 0.
Answer: T
Proof.
limn→∞
anbn = limn→∞
an limn→∞
bn
= 0
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = 0 and bn is bounded, then
limn→∞
anbn = 0.
Answer: TCaution! The previous proof does not work.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If a2n is convergent, then an is convergent.
Answer: F
Example
For an = (−1)n, a2n converges to 1 but an is divergent.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If an is convergent, then |an| is convergent.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If |an| is convergent, then an is convergent.
Answer: F
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If an and bn are divergent, then an + bn is divergent.
Answer: F
Example
The sequences an = n and bn = −n are divergent but an + bn = 0converges to 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
bn = +∞, then
limn→∞
anbn
= 0.
Answer: F
Example
For an = n2 and bn = n, we have limn→∞
bn = +∞ but
anbn
=n2
n= n is divergent.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If an is convergent and limn→∞
bn = ±∞, then
limn→∞
anbn
= 0.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If an is bounded and limn→∞
bn = ±∞, then
limn→∞
anbn
= 0.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose an is bounded. Suppose bn is a sequence and there existsN such that bn = an for any n > N . Then bn is bounded.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose limn→∞
an = a. Suppose bn is a sequence and there exists
N such that bn = an for any n > N . Then
limn→∞
bn = a.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose an and bn are convergent sequences such that an < bn forany n. Then
limn→∞
an < limn→∞
bn.
Answer: F
Example
The sequences an = 0 and bn =1
nsatisfy an < bn for any n.
Howeverlimn→∞
an 6< limn→∞
bn
because both of them are 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
Suppose an and bn are convergent sequences such that an ≤ bn forany n. Then
limn→∞
an ≤ limn→∞
bn.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
an = a, then
limn→∞
a2n = limn→∞
a2n+1 = a.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
a2n = limn→∞
a2n+1 = a, then
limn→∞
an = a.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If an is convergent, then
limn→∞
(an+1 − an) = 0.
Answer: T
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
(an+1 − an) = 0, then an is convergent.
Answer: F
Example
Let an =√n. Then lim
n→∞(an+1 − an) = 0 and an is divergent.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Exercise (True or False)
If limn→∞
(an+1 − an) = 0 and an is bounded, then an is convergent.
Answer: F
Example
0,1
2, 1,
2
3,1
3, 0,
1
4,2
4,3
4, 1,
4
5,3
5,2
5,1
5, 0,
1
6,2
6, . . .
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Let a > 0 be a positive real number.
limn→∞
an =
+∞, if a > 1
1, if a = 1
0, if 0 < a < 1
.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
limn→∞
2n− 5
3n+ 1= lim
n→∞
2− 5n
3 + 1n
=2− 0
3 + 0
=2
3
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
limn→∞
n3 − 2n+ 7
4n3 + 5n2 − 3= lim
n→∞
1− 2n2 + 7
n3
4 + 5n −
3n3
=1
4
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
limn→∞
3n−√4n2 + 1
3n+√9n2 + 1
= limn→∞
3−√4n2+1n
3 +√9n2+1n
= limn→∞
3−√
4 + 1n2
3 +√
9 + 1n2
=1
6
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
limn→∞
(n−√n2 − 4n+ 1)
= limn→∞
(n−√n2 − 4n+ 1)(n+
√n2 − 4n+ 1)
n+√n2 − 4n+ 1
= limn→∞
n2 − (n2 − 4n+ 1)
n+√n2 − 4n+ 1
= limn→∞
4n− 1
n+√n2 − 4n+ 1
= limn→∞
4− 1n
1 +√1− 4
n + 1n2
= 2
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
limn→∞
ln(n4 + 1)
ln(n3 + 1)= lim
n→∞
ln(n4(1 + 1n4 ))
ln(n3(1 + 1n3 ))
= limn→∞
lnn4 + ln(1 + 1n4 )
lnn3 + ln(1 + 1n3 )
= limn→∞
4 lnn+ ln(1 + 1n4 )
3 lnn+ ln(1 + 1n3 )
= limn→∞
4 +ln(1+ 1
n4 )
lnn
3 +ln(1+ 1
n3 )
lnn
=4
3
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Squeeze theorem)
Suppose an, bn, cn are sequences such that an ≤ bn ≤ cn for any nand lim
n→∞an = lim
n→∞cn = L. Then bn is convergent and
limn→∞
bn = L.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
If an is bounded and limn→∞
bn = 0, then limn→∞
anbn = 0.
Proof.
Since an is bounded, there exists M such that −M < an < M for any n.Thus
−M |bn| < anbn < M |bn|
for any n. Now
limn→∞
(−M |bn|) = limn→∞
M |bn| = 0.
Therefore by squeeze theorem, we have
limn→∞
anbn = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Find limn→∞
√n+ (−1)n√n− (−1)n
.
Solution
Since (−1)n is bounded and limn→∞
1√n= 0, we have
limn→∞
(−1)n√n
= 0 and therefore
limn→∞
√n+ (−1)n√n− (−1)n
= limn→∞
1 + (−1)n√n
1− (−1)n√n
= 1
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Show that limn→∞
2n
n!= 0.
Proof.
Observe that for any n ≥ 3,
0 <2n
n!= 2
(2
2· 23· 24· · · 2
n− 1
)2
n≤ 2 · 2
n=
4
n
and limn→∞
4
n= 0. By squeeze theorem, we have
limn→∞
2n
n!= 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Monotone convergence theorem)
If an is bounded and monotonic, then an is convergent.
Bounded and Monotonic⇒ Convergent
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Let an be the sequence defined by the recursive relation{an+1 =
√an + 1 for n ≥ 1
a1 = 1.
Find limn→∞
an.
n an1 1
2 1.414213562
3 1.553773974
4 1.598053182
5 1.611847754
10 1.618016542
15 1.618033940
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Solution
Suppose limn→∞
an = a. Then limn→∞
an+1 = a and thus
a =√a+ 1
a2 = a+ 1
a2 − a− 1 = 0
By solving the quadratic equation, we have
a =1 +√5
2or
1−√5
2.
It is obvious that a > 0. Therefore
a =1 +√5
2≈ 1.6180339887
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Solution
The above solution is not complete. The solution is valid only afterwe have proved that lim
n→∞an exists and is positive. This can be
done by using monotone convergent theorem. We are going toshow that an is bounded and monotonic.BoundednessWe prove that 1 ≤ an < 2 for all n ≥ 1 by induction.(Base case) When n = 1, we have a1 = 1 and 1 ≤ a1 < 2.(Induction step) Assume that 1 ≤ ak < 2. Then
ak+1 =√ak + 1 ≥
√1 + 1 > 1
ak+1 =√ak + 1 <
√2 + 1 < 2
Thus 1 ≤ an < 2 for any n ≥ 1 which implies that an is bounded.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Solution
MonotonicityWe prove that an+1 > an for any n ≥ 1 by induction.(Base case) When n = 1, a1 = 1, a2 =
√2 and thus a2 > a1.
(Induction step) Assume that
ak+1 > ak (Induction hypothesis).
Then
ak+2 =√ak+1 + 1 >
√ak + 1 (by induction hypothesis)
= ak+1
This completes the induction step and thus an is strictly increasing.We have proved that an is bounded and strictly increasing. Therefore anis convergent by monotone convergence theorem. Since an ≥ 1 for anyn, we have lim
n→∞an ≥ 1 is positive.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Let an =Fn+1
Fnwhere Fn is the Fibonacci’s sequence defined by{
Fn+2 = Fn+1 + Fn
F1 = F2 = 1.
Find limn→∞
an.
n an1 12 23 1.54 1.6666666665 1.610 1.61818181815 1.61803278720 1.618033999
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
For any n ≥ 1,
1 Fn+2Fn − F 2n+1 = (−1)n+1
2 Fn+3Fn − Fn+2Fn+1 = (−1)n+1
Proof
1 When n = 1, we have F3F1 − F 22 = 2 · 1− 12 = 1 = (−1)2. Assume
Fk+2Fk − F 2k+1 = (−1)k+1.
Then
Fk+3Fk+1 − F 2k+2 = (Fk+2 + Fk+1)Fk+1 − F 2
k+2
= Fk+2(Fk+1 − Fk+2) + F 2k+1
= −Fk+2Fk + F 2k+1
= (−1)k+2 (by induction hypothesis)
Therefore Fn+2Fn − F 2n+1 = (−1)n+1 for any n ≥ 1.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof.
The proof for the second statement is basically the same. Whenn = 1, we have F4F1 − F3F2 = 3 · 1− 2 · 1 = 1 = (−1)2. Assume
Fk+3Fk − Fk+2Fk+1 = (−1)k+1.
Then
Fk+4Fk+1 − Fk+3Fk+2 = (Fk+3 + Fk+2)Fk+1 − Fk+3Fk+2
= Fk+3(Fk+1 − Fk+2) + Fk+2Fk+1
= −Fk+3Fk + Fk+2Fk+1
= −(−1)k+1 (by induction hypothesis)
= (−1)k+2
Therefore Fn+3Fn − Fn+2Fn+1 = (−1)n+1 for any n ≥ 1.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let an =Fn+1
Fn.
1 The sequence a1, a3, a5, a7, · · · , is strictly increasing.
2 The sequence a2, a4, a6, a8, · · · , is strictly decreasing.
Proof.
For any k ≥ 1, we have
a2k+1 − a2k−1 =F2k+2
F2k+1− F2k
F2k−1=F2k+2F2k−1 − F2k+1F2k
F2k+1F2k−1
=(−1)2k
F2k+1F2k−1=
1
F2k+1F2k−1> 0
Therefore a1, a3, a5, a7, · · · , is strictly increasing. The second statementcan be proved in a similar way.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
limk→∞
(a2k+1 − a2k) = 0
Proof.
For any k ≥ 1,
a2k+1 − a2k =F2k+2
F2k+1− F2k+1
F2k
=F2k+2F2k − F 2
2k+1
F2k+1F2k=
1
F2k+1F2k
Therefore
limk→∞
(a2k+1 − a2k) = limk→∞
1
F2k+1F2k= 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
limn→∞
Fn+1
Fn=
1 +√5
2
Proof
First we prove that an = Fn+1
Fnis convergent.
an is bounded. (1 ≤ an ≤ 2 for any n.)a2k+1 and a2k are convergent. (They are bounded and monotonic.)
limk→∞
(a2k+1 − a2k) = 0⇒ limk→∞
a2k+1 = limk→∞
a2k
It follows that an is convergent and
limn→∞
an = limk→∞
a2k+1 = limk→∞
a2k.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof.
To evaluate the limit, suppose limn→∞
Fn+1
Fn= L. Then
L = limn→∞
Fn+2
Fn+1= lim
n→∞
Fn+1 + Fn
Fn+1= lim
n→∞
(1 +
Fn
Fn+1
)= 1 +
1
L
L2 − L− 1 = 0
By solving the quadratic equation, we have
L =1 +√5
2or
1−√5
2.
We must have L ≥ 1 since an ≥ 1 for any n. Therefore
L =1 +√5
2.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Remarks
The limit can be calculate directly using the formula
Fn =αn − βn
α− β
=1√5
((1 +√5
2
)n
−
(1−√5
2
)n)
where
α =1 +√5
2, β =
1−√5
2
are the roots of the quadratic equation
x2 − x− 1 = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let
an =
(1 +
1
n
)n
bn =
n∑k=0
1
k!= 1 + 1 +
1
2!+
1
3!+ · · ·+ 1
n!
Then
1 an < bn for any n > 1.
2 an and bn are convergent and
limn→∞
an = limn→∞
bn
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
an =
(1 +
1
n
)n
bn =
n∑k=0
1
k!= 1 + 1 +
1
2!+
1
3!+ · · ·+ 1
n!
n an bn1 2 25 2.48832 2.71666666666610 2.593742 2.718281801146100 2.704813 2.718281828459
100000 2.718268 2.718281828459
The limit of the two sequences is the important Euler’s number
e ≈ 2.71828 18284 59045 23536 . . . .
which is also known as the Napier’s constant.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Convergence of infinite series)
We say that an infinite series
∞∑k=1
ak = a1 + a2 + a3 + · · ·
is convergent if the sequence of partial sums
sn =n∑
k=1
ak = a1 + a2 + a3 + · · ·+ an is convergent. If the infinite series is
convergent, then we define
∞∑k=1
ak = limn→∞
sn = limn→∞
n∑k=1
ak.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Function)
A real valued function on a subset D ⊂ R is a real value f(x)assigned to each of the values x ∈ D. The set D is called thedomain of the function.
Given an expression f(x) in x, the domain D is understood to betaken as the set of all real numbers x such that f(x) is defined.This is called the maximum domain of definition of f(x).
Definition (Graph of function)
Let f(x) is a real valued function. The graph of f(x) is the set
{(x, y) ∈ R2 : y = f(x)}.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition
Let f(x) be a real valued function and D be its domain. We saythat f(x) is
1 injective if for any x1, x2 ∈ D with x1 6= x2, we havef(x1) 6= f(x2).
2 surjective if for any real number y ∈ R, there exists x ∈ Dsuch that f(x) = y.
3 bijective if f(x) is both injective and surjective.
Definition
Let f(x) be a real valued function. We say that f(x) is
1 even if f(−x) = f(x) for any x.
2 odd if f(−x) = −f(x) for any x.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
f(x) Domain Injective Surjective Bijective Even Odd
2x− 3 R X X X × ×x3 − 2x2 R × X × × ×
1
xx 6= 0 X × × × X
4x
x2 + 1R × × × × X
x
x2 − 1x 6= ±1 × X × × X
x2 − 1
x2x 6= 0 × X × X ×
√4− x2 −2 ≤ x ≤ 2 × × × X ×1√x+ 4
x > −4 X × × × ×
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Limit of function)
Let f(x) be a real valued function.
1 We say that a real number l is a limit of f(x) at x = a if for any ε > 0,there exists δ > 0 such that
if 0 < |x− a| < δ, then |f(x)− l| < ε
and writelimx→a
f(x) = l.
2 We say that a real number l is a limit of f(x) at +∞ if for any ε > 0,there exists R > 0 such that
if x > R, then |f(x)− l| < ε
and writelim
x→+∞f(x) = l.
The limit of f(x) at −∞ is defined similarly.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
1 Note that for the limit of f(x) at x = a to exist, f(x) maynot be defined at x = a and even if f(a) is defined, the valueof f(a) does not affect the value of lim
x→af(x).
2 The limit of f(x) at x = a may not exist. However the limit isunique if it exists.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
limx→a
f(x) = l
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
limx→a
f(x)
does not exist
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Sequential criterion for limits of functions)
Let f(x) be a real valued function. Then
limx→a
f(x) = l
if and only if for any sequence xn of real numbers with limn→∞
xn = a, we
havelimn→∞
f(xn) = l.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let f(x), g(x) be functions such that limx→a
f(x), limx→a
g(x) exist and
c be a real number. Then
1 limx→a
(f(x) + g(x)) = limx→a
f(x) + limx→a
g(x)
2 limx→a
cf(x) = c limx→a
f(x)
3 limx→a
f(x)g(x) = limx→a
f(x) limx→a
g(x)
4 limx→a
f(x)
g(x)=
limx→a
f(x)
limx→a
g(x)if lim
x→ag(x) 6= 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let f(u) be a function of u and u = g(x) be a function of x.Suppose
1 limx→a
g(x) = b ∈ [−∞,+∞]
2 limu→b
f(u) = l
3 g(x) 6= b when x 6= a or f(b) = l.
Thenlimx→a
f ◦ g(x) = l.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
1. limx→+∞
6x3 + 2x2 − 5
2x3 − 3x+ 1= lim
x→+∞
6 + 2x −
5x3
2− 3x2 + 1
x3
= limy→0
6 + 2y − 5y3
2− 3y + y3
= 3
2. limn→∞
(1 +
1
n2
)n2
= limm→∞
(1 +
1
m
)m
= e
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Squeeze theorem)
Let f(x), g(x), h(x) be real valued functions. Suppose
1 f(x) ≤ g(x) ≤ h(x) for any x 6= a on a neighborhood of a, and
2 limx→a
f(x) = limx→a
h(x) = l.
Then the limit of g(x) at x = a exists and limx→a
g(x) = l.
Theorem
Suppose
1 f(x) is bounded, and
2 limx→a
g(x) = 0
Then limx→a
f(x)g(x) = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Exponential function)
The exponential function is defined for real number x ∈ R by
ex = limn→∞
(1 +
x
n
)n= 1 + x+
x2
2!+x3
3!+x4
4!+ · · ·
1 It can be proved that the two limits in the definition exist andconverge to the same value for any real number x.
2 ex is just a notation for the exponential function. One shouldnot interpret it as ‘e to the power x’.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
For any x, y ∈ R, we have
ex+y = exey.
Caution! One cannot use law of indices to prove the above identity.It is because ex is just a notation for the exponential function andit does not mean ‘e to the power x’. In fact we have not definedwhat ax means when x is a real number which is not rational.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
1 ex > 0 for any real number x.
2 ex is strictly increasing.
Proof.
1 For any x > 0, we have ex > 1 + x > 1. If x < 0, then
exe−x = ex+(−x) = e0 = 1
ex =1
e−x> 0
since e−x > 1. Therefore ex > 0 for any x ∈ R.
2 Let x, y be real numbers with x < y. Then y − x > 0 which impliesey−x > 1. Therefore
ey = ex+(y−x) = exey−x > ex.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Logarithmic function)
The logarithmic function is the function ln : R+ → R defined forx > 0 by
y = lnx if ey = x.
In other words, lnx is the inverse function of ex.
It can be proved that for any x > 0, there exists unique realnumber y such that ey = x.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
1 lnxy = lnx+ ln y
2 lnx
y= lnx− ln y
3 lnxn = n lnx for any integer n ∈ Z.
Proof.
1 Let u = lnx and v = ln y. Then x = eu, y = ev and we have
xy = euev = eu+v = elnx+ln y
which means lnxy = lnx+ ln y.
Other parts can be proved similarly.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Cosine and sine functions)
The cosine and sine functions are defined for real number x ∈ Rby the infinite series
cosx = 1− x2
2!+x4
4!− x6
6!+ · · ·
sinx = x− x3
3!+x5
5!− x7
7!+ · · ·
1 When the sine and cosine are interpreted as trigonometricratios, the angles are measured in radian. (1800 = π)
2 The series for cosine and sine are convergent for any realnumber x ∈ R.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
There are four more trigonometric functions namely tangent, cotangent,secant and cosecant functions. All of them are defined in terms of sineand cosine.
Definition (Trigonometric functions)
tanx =sinx
cosx, for x 6= 2k + 1
2π, k ∈ Z
cotx =cosx
sinx, for x 6= kπ, k ∈ Z
secx =1
cosx, for x 6= 2k + 1
2π, k ∈ Z
cscx =1
sinx, for x 6= kπ, k ∈ Z
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Trigonometric identities)
1 cos2 x+ sin2 x = 1; sec2 x− tan2 x = 1; csc2 x− cot2 x = 1
2 cos(x± y) = cosx cos y ∓ sinx sin y;
sin(x± y) = sinx cos y ± cosx sin y;
tan(x± y) =tanx± tan y
1∓ tanx tan y
3 cos 2x = cos2 x− sin2 x = 2 cos2 x− 1 = 1− 2 sin2 x;
sin 2x = 2 sinx cosx;
tan 2x =2 tanx
1− tan2 x
4 2 cosx cos y = cos(x+ y) + cos(x− y)
2 cosx sin y = sin(x+ y)− sin(x− y)2 sinx sin y = cos(x− y)− cos(x+ y)
5 cosx+ cos y = 2 cos(
x+y2
)cos(
x−y2
)cosx− cos y = −2 sin
(x+y2
)sin(
x−y2
)sinx+ sin y = 2 sin
(x+y2
)cos(
x−y2
)sinx− sin y = 2 cos
(x+y2
)sin(
x−y2
)MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Hyperbolic function)
The hyperbolic functions are defined for x ∈ R by
coshx =ex + e−x
2= 1 +
x2
2!+x4
4!+x6
6!+ · · ·
sinhx =ex − e−x
2= x+
x3
3!+x5
5!+x7
7!+ · · ·
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Hyperbolic identities)
1 cosh2 x− sinh2 x = 1
2 cosh(x+ y) = coshx cosh y + sinhx sinh ysinh(x+ y) = sinhx cosh y + coshx sinh y
3 cosh 2x = cosh2 x+ sinh2 x = 2 cosh2 x− 1 = 1 + 2 sinh2 x;
sinh 2x = 2 sinhx coshx
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
1 limx→0
ex − 1
x= 1
2 limx→0
ln(1 + x)
x= 1
3 limx→0
sinx
x= 1
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof. limx→0
ex − 1
x= 1.
For any −1 < x < 1 with x 6= 0, we have
ex − 1
x= 1 +
x
2!+x2
3!+x3
4!+x4
5!+ · · ·
≤ 1 +x
2+
(x2
4+x2
8+x2
16+ · · ·
)= 1 +
x
2+x2
2
ex − 1
x= 1 +
x
2!+x2
3!+x3
4!+ · · ·
≥ 1 +x
2−(x2
4+x2
8+x2
16+ · · ·
)= 1 +
x
2− x2
2
and limx→0
(1+x
2+x2
2) = lim
x→0(1+
x
2− x2
2) = 1. Therefore lim
x→0
ex − 1
x= 1.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Figure: limx→0
ex − 1
x= 1
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof. limx→0
ln(1 + x)
x= 1.
Let y = ln(1 + x). Then
ey = 1 + x
x = ey − 1
and x→ 0 as y → 0. We have
limx→0
ln(1 + x)
x= lim
y→0
y
ey − 1
= 1
Note that the first part implies limy→0
(ey − 1) = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof. limx→0
sinx
x= 1.
Note thatsinx
x= 1− x2
3!+x4
5!− x6
7!+x8
9!− x10
11!+ · · · .
For any −1 < x < 1 with x 6= 0, we have
sinx
x= 1−
(x2
3!− x4
5!
)−(x6
7!− x8
9!
)− · · · ≤ 1
sinx
x= 1− x2
6+
(x4
5!− x6
7!
)+
(x8
9!− x10
11!
)+ · · · ≥ 1− x2
6
and limx→0
1 = limx→0
(1− x2
6) = 1. Therefore
limx→0
sinx
x= 1.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Figure: limx→0
sinx
x= 1
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let k be a positive integer.
1 limx→+∞
xk
ex= 0
2 limx→+∞
(lnx)k
x= 0
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Proof.
1 For any x > 0,
ex = 1 + x+x2
2!+
x3
3!+ · · · >
xk+1
(k + 1)!
and thus
0 <xk
ex<
(k + 1)!
x.
Moreover limx→∞
(k + 1)!
x= 0. Therefore
limx→+∞
xk
ex= 0.
2 Let x = ey . Then x→ +∞ as y → +∞ and lnx = y. We have
limx→+∞
(lnx)k
x= lim
y→+∞
yk
ey= 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
1. limx→4
x2 − 16√x− 2
= limx→4
(x− 4)(x+ 4)(√x+ 2)
(√x− 2)(
√x+ 2)
= limx→4
(x− 4)(x+ 4)(√x+ 2)
x− 4
= limx→4
(x+ 4)(√x+ 2) = 32
2. limx→+∞
3e2x + ex − x4
4e2x − 5ex + 2x4= lim
x→+∞
3 + e−x − x4e−2x
4− 5e−x + 2x4e−2x=
3
4
3. limx→+∞
ln(2e4x + x3)
ln(3e2x + 4x5)= lim
x→+∞
4x+ ln(2 + x3e−4x)
2x+ ln(3 + 4x5e−2x)
= limx→+∞
4 +ln(2+x3e−4x)
x
2 +ln(3+4x5e−2x)
x
= 2
4. limx→−∞
(x+√x2 − 2x) = lim
x→−∞
(x+√x2 − 2x)(x−
√x2 − 2x)
x−√x2 − 2x
= limx→−∞
2x
x−√x2 − 2x
= limx→−∞
2
1 +√
1− 2x
= 1
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
5. limx→0
sin 6x− sinx
sin 4x− sin 3x= lim
x→0
6 sin 6x6x
− sin xx
4 sin 4x4x
− 3 sin 3x3x
=6− 1
4− 3= 5
6. limx→0
1− cosx
x tanx= lim
x→0
(1− cosx)(1 + cosx)
x sin xcos x
(1 + cosx)
= limx→0
(1− cos2 x) cosx
x sinx(1 + cosx)
= limx→0
(sinx
x
)cosx
1 + cosx=
1
2
7. limx→0
e2x − 1
ln(1 + 3x)= lim
x→0
2
3·e2x − 1
2x·
3x
ln(1 + 3x)=
2
3
8. limx→0
x ln(1 + sinx)
1−√cosx
= limx→0
x(1 +√cosx)(1 + cosx) ln(1 + sinx)
1− cos2 x
= limx→0
x
sinx·ln(1 + sinx)
sinx(1 +
√cosx)(1 + cosx)
= 4
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition (Continuity)
Let f(x) be a real valued function. We say that f(x) is continuous atx = a if
limx→a
f(x) = f(a).
In other words, f(x) is continuous at x = a if for any ε > 0, there existsδ > 0 such that
if |x− a| < δ, then |f(x)− f(a)| < ε.
We say that f(x) is continuous on an interval in R if f(x) is continuousat every point on the interval.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Let g(u) be a function in u and u = f(x) be a function in x.Suppose g(u) is continuous and the limit of f(x) at x = a exists.Then
limx→a
(g ◦ f)(x)) = limx→a
g(f(x)) = g(limx→a
f(x)).
xf−−→ u = f(x)
g−−→ (g ◦ f)(x) = g(u) = g(f(x))
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
1 For any non-negative integer n, f(x) = xn is continuous on R.2 The functions ex, cosx, sinx are continuous on R.3 The logarithmic function lnx is continuous on R+.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
Suppose f(x), g(x) are continuous functions and c is a realnumber. Then the following functions are continuous.
1 f(x) + g(x)
2 cf(x)
3 f(x)g(x)
4f(x)
g(x)at the points where g(x) 6= 0.
5 f ◦ g(x)
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Definition
The absolute value of x ∈ R is defined by
|x| =
{−x, if x < 0
x, if x ≥ 0
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example (Piecewise defined function)
a 1 5
limx→a−
f(x) 3 2
limx→a+
f(x) 0 2
limx→a
f(x) does not exist 2
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem
A function f(x) is continuous at x = a if
limx→a+
f(x) = limx→a−
f(x) = f(a).
The theorem is usually used to check whether a piecewise definedfunction is continuous.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
The following functions are not continuous at x = a.
limx→a−
f(x) does not exist limx→a
f(x) does not exist
limx→a−
f(x) 6= limx→a+
f(x) limx→a
f(x) 6= f(a)
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Given that the function
f(x) =
2x− 1 if x < 2
a if x = 2
x2 + b if x > 2
is continuous at x = 2. Find the value of a and b.
Solution
Note that
limx→2−
f(x) = limx→2−
(2x− 1) = 3
limx→2+
f(x) = limx→2+
(x2 + b) = 4 + b
f(2) = a
Since f(x) is continuous at x = 2, we have 3 = 4 + b = a which implies a = 3and b = −1.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Example
Prove that the function
f(x) =
sin
(1
x
), if x 6= 0
0, if x = 0
is not continuous at x = 0.
Proof.
Let xn =2
(2n+ 1)πfor n = 1, 2, 3, . . . . Then lim
n→∞xn = 0 and
f(xn) = sin
((2n+ 1)π
2
)= (−1)n.
Thus limn→∞
f(xn) does not exist. Therefore f(x) is not continuous at
x = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
f(x) is not continuous at x = 0.
MATH1010 University Mathematics
LimitsSequencesLimits and Continuity
Theorem (Intermediate value theorem)
Suppose f(x) is a function which is continuous on a closed andbounded interval [a, b]. Then for any real number η between f(a)and f(b), there exists ξ ∈ (a, b) such that f(ξ) = η.
Theorem (Extreme value theorem)
Suppose f(x) is a function which is continuous on a closed andbounded interval [a, b]. Then there exists α, β ∈ [a, b] such thatfor any x ∈ [a, b], we have
f(α) ≤ f(x) ≤ f(β).
MATH1010 University Mathematics