Benginning Calculus Lecture notes 2 - limits and continuity

Beginning Calculus
- Limits and Continuity -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
(D1) Limits and Continuity 1 / 54

The Limit of a Function Limits of Trigonometric Functions In…nite Limits and Limits at In…nity Continuity
Learning Outcomes
Determine the existence of limits of functions
Compute the limits of functions
Determine the continuity of functions.
Connect the idea of limits and continuity of functions.

Limits
De…nition 1
The limit of f (x), as x approaches a, equals L, denoted by
lim
x!a
f (x) = L or f (x) ! L as x ! a (1)
if the values of f (x) moves arbitrarily close to L as x moves su¢ ciently
close to a (on either side of a ) but not equal to a.

Example
lim
x!2
x2 x + 2 = 4
0 2 4
0
5
10
x
y x < 2 f (x) x > 2 f (x)
1.0 2.000000 3.0 8.000000
1.5 2.750000 2.5 5.750000
1.9 3.710000 2.1 4.310000
1.99 3.970100 2.01 4.030100
1.995 3.985025 2.005 4.015025
1.999 3.997001 2.001 4.003001

Example
Estimate the value of lim
t!0
p
t2 + 9 3
t2
.
f
0
B
B
B
B
B
B
B
B
B
B
B
B
@
t
0.1
0.001
0.0001
0.00001
0.00001
0.0001
0.001
0.1
1
C
C
C
C
C
C
C
C
C
C
C
C
A
=
0
B
B
B
B
B
B
B
B
B
B
B
B
B
@
1
t2
p
t2 + 9 3
0.166 62
0.166 67
0.166 67
0.166 67
0.166 67
0.166 67
0.166 67
0.166 62
1
C
C
C
C
C
C
C
C
C
C
C
C
C
A

Example - continue
-4 -2 0 2 4
0.12
0.13
0.14
0.15
0.16
lim
t!0
p
t2 + 9 3
t2
=
1
6

Example
f (x) = x + 1.
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y
lim
x!2
f (x) = 3

Example
g (x) =
x + 1 if x 2
(x 2)2
+ 3 if x > 2
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y
lim
x!2
g (x) = 3

Example
h (x) =
x + 1 if x < 2
(x 2)2
+ 3 if x > 2
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y
lim
x!2
h (x) = 3, eventhough h is not de…ned at x = 2.

One-Sided Limits
Left-hand limit of f
lim
x!a
f (x) = L (2)
Right-hand limit of f
lim
x!a+
f (x) = L (3)
lim
x!a
f (x) = L , f lim
x!a
f (x) = lim
x!a+
f (x) = L. (4)

Example
f (x) =
x + 1 if x 2
(x 2)2
+ 1 if x > 2
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y
lim
x!2
f (x) = 3 and lim
x!2+
f (x) = 1
lim
x!2
f (x) does not exist (DNE), eventhough f is de…ned at x = 2.

Example
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y
Find:
f (2) and f (4)
lim
x!2
f (x) , lim
x!2+
f (x) , lim
x!2
f (x)
lim
x!4
f (x) , lim
x!4+
f (x) lim
x!4
f (x)

Properties of Limits
Suppose that lim
x!a
f (x) and lim
x!a
g (x) exists. Then,
1. lim
x!a
(cf (x)) = c lim
x!a
f (x) , for any constant c
2. lim
x!a
[f (x) g (x)] = lim
x!a
f (x) lim
x!a
g (x)
3. lim
x!a
[f (x) g (x)] =
h
lim
x!a
f (x)
i h
lim
x!a
g (x)
i
4. lim
x!a
f (x)
g (x)
=
lim
x!a
f (x)
lim
x!a
g (x)
provided that lim
x!a
g (x) 6= 0

Properties of Limits - continue
5. lim
x!a
x = a
6. lim
x!a
c = c, for any constant c.
7. lim
x!a
[f (x)]n
=
h
lim
x!a
f (x)
in
where n 2 Z+.
8. lim
x!a
n
p
x = n
p
a where n 2 Z+ (If n is even, we assume that a > 0 ).
9. lim
x!a
n
p
f (x) = n
q
lim
x!a
f (x) where n 2 Z+. (If n is even, we assume
that lim
x!a
f (x) > 0 ).

Direct Substitution Property
If f is a polynomial or a rational function and a is in the domain of f ,
then
lim
x!a
f (x) = f (a) (5)

Example
lim
x!5
2x2
3x + 4 = lim
x!5
2x2
lim
x!5
3x + lim
x!5
4
= 2 lim
x!5
x2
3 lim
x!5
x + lim
x!5
4
= 2 52
3 (5) + 4
= 39

Example
lim
x! 2
x3 + 2x2 1
5 3x
=
lim
x! 2
x3 + 2x2 1
lim
x! 2
(5 3x)
=
lim
x! 2
x3 + 2 lim
x! 2
x2 lim
x! 2
1
lim
x! 2
5 3 lim
x! 2
x
=
( 2)3
+ 2 ( 2)2
1
5 3 ( 2)
=
1
11

De…nition 2
If f (x) = g (x) when x 6= a, then lim
x!a
f (x) = lim
x!a
g (x) , provided that
the limits exist.

Example
lim
x!1
x2 1
x 1
. For x 6= 1,
x2 1
x 1
=
(x 1) (x + 1)
x 1
= x + 1
lim
x!1
x2 1
x 1
= lim
x!1
x + 1 = 2
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y

Example
lim
h!0
(3 + h)2
9
h
. For h 6= 0,
(3 + h)2
9
h
=
9 + 6h + h2 9
h
= 6 + h
lim
h!0
(3 + h)2
9
h
= lim
h!0
6 + h = 6

Example
lim
x!2
jx 2j
x 2
.
For x 2 > 0, jx 2j = x 2.
lim
x!2
jx 2j
x 2
= lim
x!2
x 2
x 2
= lim
x!2
1 = 1
For x 2 < 0, jx 2j = (x 2) = 2 x.
lim
x!2
jx 2j
x 2
= lim
x!2
(x 2)
x 2
= lim
x!2
1 = 1
lim
x!2
jx 2j
x 2
DNE

Remark 1
lim
θ!0
cos θ 1
θ
= 0
Rewrite:
1 cos θ
θ
to make the numerator stays positive.
θ
1
O
A
BC
BC = 1 cos θ, arclength AB = θ.
1 cos θ
θ
! 0 as θ ! 0

Remark 2
lim
θ!0
sin θ
θ
= 1
θ
1
O
A
BC
AC = sin θ, arclength AB = θ
sin θ
θ
! 1 as θ ! 0.
Principle: Short pieces of curves are nearly straight.

Example
lim
θ!0
tan θ
θ
tan θ
θ
=
sin θ
cos θ
θ
=
sin θ
θ cos θ
=
sin θ
θ
1
cos θ
lim
θ!0
tan θ
θ
= lim
θ!0
sin θ
θ
lim
θ!0
1
cos θ
= 1 1 = 1

Example
lim
θ!0
sin 2θ
tan θ
sin 2θ
tan θ
=
sin 2θ
θ
tan θ
θ
=
2 sin 2θ
2θ
tan θ
θ
lim
θ!0
sin 2θ
tan θ
= lim
θ!0
2 sin 2θ
2θ
tan θ
θ
=
lim
θ!0
2 sin 2θ
2θ
lim
θ!0
tan θ
θ
=
2
1
= 2

In…nite Limits
De…nition 3
Let f de…ned on both sides of a, except possibly at a itself. Then
lim
x!a
f (x) = ∞ or lim
x!a
f (x) = ∞ (6)
means that the values of f (x) can be made arbitrarily large (as large as
possible) by taking x su¢ ciently close to a, but not equal to a. x = a is
the vertical asymptote.
y
x
y = f(x)
x = a
a
y
x
y = f(x)
x = a
a

Example
lim
x!3+
2x
x 3
= +∞ and lim
x!3
2x
x 3
= ∞
-5 5 10
-5
5
10
x
y
x = 3
The vertical asymptote is at x = 3.

Example
f (x) = tan x =
sin x
cos x
The vertical asymptote can be obtained by setting cos x = 0, that is,
x =
π
2
x = (2n + 1)
π
2
, n 2 Z
x
y

Limits at In…nity
De…nition 4 (Limits at In…nity)
(a) Let f be a function de…ned on some interval (a, ∞) . Then
lim
x!∞
f (x) = L (7)
means that the values of f (x) can be made arbitrarily close to L by
taking x su¢ ciently large.
(b) Let f be a function de…ned on some interval ( ∞, a) . Then
lim
x! ∞
f (x) = L (8)
means that the values of f (x) can be made arbitrarily close to L by
taking x su¢ ciently large negative.

Horizontal Asymptotes
The line y = L is called a horizontal asymptote of the curve y = f (x) if
either
lim
x!∞
f (x) = L or lim
x! ∞
f (x) = L (9)

Example
f (x) =
x2 1
x2 + 1
lim
x!∞
f (x) = 1 = lim
x! ∞
f (x)
-10 -5 5 10
-1
1
2
x
y
No vertical asymtote.
The horizontal asymptote is y = 1.

Example
f (x) =
1
x
.
lim
x!0
1
x
= ∞, lim
x!0+
1
x
= +∞
lim
x!∞
1
x
= 0 = lim
x! ∞
1
x
Vertical asymtote at x = 0
The horizontal asymptote at y = 0.
-4 -2 2 4
-4
-2
2
4
x
y

Example - Finding the Asymptotes
f (x) =
3x2 x 2
5x2 + 4x + 1
lim
x!∞
3x2 x 2
5x2 + 4x + 1
= lim
x!∞
3x2
x2
x
x2
2
x2
5x2
x2
+
4x
x2
+
1
x2
= lim
x!∞
3
1
x
2
x2
5 +
4
x
+
1
x2
=
lim
x!∞
3
1
x
2
x2
lim
x!∞
5 +
4
x
+
1
x2
=
lim
x!∞
3 lim
x!∞
1
x
lim
x!∞
2
x2
lim
x!∞
5 + lim
x!∞
4
x
+ lim
x!∞
1
x2
=
3 0 0
5 + 0 + 0
=
3
5
The horizontal asymptote is y =
3
5
.

f (x) =
p
2x2 + 1
3x 5
.
lim
x!∞
p
2x2 + 1
3x 5
= lim
x!∞
p
2x2 + 1
p
x2
3x 5
x
,
p
x2 = x for x > 0
= lim
x!∞
r
2x2
x2
+
1
x2
3x
x
5
x
= lim
x!∞
r
2 +
1
x2
3
5
x
=
lim
x!∞
r
2 +
1
x2
lim
x!∞
3
5
x
=
r
lim
x!∞
2 + lim
x!∞
1
x2
lim
x!∞
3 lim
x!∞
5
x
=
p
2
3

Example - Finding the Asymptotes - continue
lim
x! ∞
p
2x2 + 1
3x 5
= lim
x! ∞
r
2 +
1
x2
3
5
x
,
p
x2 = x for x < 0
=
lim
x!∞
r
2 +
1
x2
lim
x! ∞
3
5
x
=
p
2
3

-4 -2 2 4
-4
-2
2
4
x
y
The horizontal asymptotes are: y =
p
2
3
.
The vertical asymptote is when 3x 5 = 0, that is, x =
5
3
.

f (x) =
p
x2 + 1 x
lim
x!∞
p
x2 + 1 x = lim
x!∞
p
x2 + 1 x
p
x2 + 1 + x
p
x2 + 1 + x
= lim
x!∞
x2 + 1 x2
p
x2 + 1 + x
= lim
x!∞
1
p
x2 + 1 + x
= lim
x!∞
1
xp
x2 + 1 + x
p
x2
= lim
x!∞
1
xr
x2
x2
+
1
x2
+ 1
= lim
x!∞
1
xr
1 +
1
x2
+ 1
=
0
p
1 + 0 + 1
= 0

-4 -2 0 2 4
5
10
x
y
The horizontal asymptote is y = 0.

Example
lim
x!∞
x3 = ∞ and lim
x! ∞
x3 = ∞.
-4 -2 2 4
-100
-50
50
100
x
y

Example
lim
x!∞
x2 x . Note that the properties of limits cannot be applied to
in…nite limits since ∞ is not a number. So,
lim
x!∞
x2
x = lim
x!∞
x (x 1) = ∞

Example
lim
x!∞
x2 + x
3 x
.
lim
x!∞
x2 + x
3 x
= lim
x!∞
x2
x
+
x
x
3
x
x
x
= lim
x!∞
x + 1
3
x
1
=
∞
1
= ∞

Continuous Functions at a Point
De…nition 5
A function f is continuous at a if
lim
x!a
f (x) = f (a) (10)
y
x
y = f(x)
a
f(a)
f (a) is de…ned (a is in the domain of f )
lim
x!a
f (x) exists.
lim
x!a
f (x) = f (a)

Example
y
x1 3 50 2 4 6
Discontinuities at 1, 3, and 5.
at a = 1, f is unde…ned
at a = 3, f is de…ned but lim
x!3
f (x) DNE;
at a = 5, f is de…ned and lim
x!5
f (x) exists, but lim
x!5
f (x) 6= f (5) .

Example
f (x) =
x2 x 2
x 2
is discontinuous at 2 because f (2) is unde…ned.
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y

Example
g (x) =
( 1
x2
if x 6= 0
1 if x = 0
is de…ned at 0 but lim
x!0
g (x) = lim
x!0
1
x2
does not exist. This discontinuity is called in…nite discontinuity.
-4 -2 2 4
-1
1
2
3
4
5
x
y

Example
h (x) =
8
<
:
x2 x 2
x 2
if x 6= 2
1 if x = 2
is de…ned at 2 and lim
x!2
h (x) = 3,
but lim
x!2
h (x) 6= h (2) . This discontinuity is called removable
discontinuity.
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y

Example
k (x) = bxc has discontinuities at all of the integers because lim
x!n
k (x)
does not exist if n is an integer. These discontinuities are called jump
discontinuities.
-1 1 2 3 4 5
-1
1
2
3
4
5
x
y

Theorem 6
If f and g are continuous at x = a and c is a constant, then the
following functions are also continuous at a.
(a) f g
(b) cf
(c) fg
(d)
f
g
if g (a) 6= 0

Theorem 7
The following functions are continuous at every number in their domains.
(a) Polynomial functions.
(b) Rational functions.
(c) Power and root functions
(d) Trigonometric Functions

Example
f (x) = x100 2x37 + 75 is a polynomial function. So it is
continuous everywhere: ( ∞, ∞)
g (x) =
x2 + 2x + 17
x2 1
is a rational function, and continuous on its
domain fx j x 6= 1g = ( ∞, 1) [ ( 1, 1) [ (1, ∞) .

Example
h (x) =
p
x +
x + 1
x 1
x + 1
x2 + 1
Let h1 (x) =
p
x; h2 (x) =
x + 1
x 1
; and h3 (x) =
x + 1
x2 + 1
.
h1 (x) is a root function and continuous on [0, ∞).
h2 (x) is a rational function and continuous on ( ∞, 1) [ (1, ∞) ,
and
h3 (x) is also a rational function and continuous everywhere on R.
So, h (x) is continuous on [0, 1) [ (1, ∞) .

Example
f (x) =
sin x
2 + cos x
Let f1 (x) = sin x, and let f2 (x) = 2 + cos x.
f1 (x) and f2 (x) are trigonometric functions. So, they are
continuous. Note that cos x 1. So, f2 (x) = 2 cos x is always
positive.
Hence, f (x) =
f1 (x)
f2 (x)
=
sin x
2 + cos x
is continuous everywhere on R.

Theorem 8
If g is continuous at a and f is continuous at g (a) , then
(f g) (x) = f (g (x)) is continuous at a.

Example
f (x) = sin x2
Let F (x) = sin x, and let G (x) = x2.
F and G are continuous on R.
So, f (x) = F (G (x)) = sin x2 is continuous on R.

Benginning Calculus Lecture notes 2 - limits and continuity

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