Continuity
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
3
Calculus I
Chapter 1
Continuity
3.1
Introduction
2
3.2
Limit of a Function
2
3.3
Properties of Limit of a Function
9
3.4
Two Important Limits
10
3.5
Left and Right Hand Limits
12
3.6
Continuous Functions
13
3.7
Properties of Continuous Functions
16
Prepared by K. F. Ngai
Page
1
Continuity
3.1
Advanced Level Pure Mathematics
Introduction
Example f ( x )  x 2 is continuous on R.
Example
y
f ( x) 
y
y  x2
0
3.2
1
is discontinuous at x = 0.
x
y
x
0
1
x
x
Limit of a Function
A. Limit of a Function at Infinity
Let f(x) be a function defined on R. lim f ( x)   means that for any  > 0 , there exists X > 0
Defintion
x 
such that when x > X ,
f (x)     .
y
y
l+
l
l
l+
l
0
N.B. (1)
X
x
0
X
x
lim f ( x)   means that the difference between f(x) and A can be made arbitrarily small when x is
x 
sufficiently large.
(2) lim f ( x)   means f(x) → A as x → ∞.
x 
(3) Infinity, ∞, is a symbol but not a real value.
There are three cases for the limit of a function when x → ∞,
1. f(x) may have a
finite limit value.
y
2. f(x) may approach
to infinity;
y
3. f(x) may oscillate or
infinitely and limit
value does not exist.
y
L
0
0
x
0
x
x
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2
Continuity
Theorem
Advanced Level Pure Mathematics
UNIQUENESS of Limit Value
If lim f ( x)  a and lim f ( x)  b , then a  b .
x 
x
Theorem
Rules of Operations on Limits
If lim f ( x) and lim g ( x ) exist , then
x
x
(a)
lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)
(b)
lim f ( x) g ( x)  lim f ( x)  lim g ( x)
(c)
lim
x 
x 
x 
x 
x 
x 
x 
f ( x)
f ( x) lim
 x 
g ( x) lim g ( x)
if lim g ( x)  0 .
x
x 
(d) For any constant k, lim [kf ( x)]  k lim f ( x) .
x 
x 
(e) For any positive integer n,
(i)
lim [ f ( x)]n  [ lim f ( x)]n
x 
(ii) lim
x 
x 
n
f ( x)  n lim f ( x)
x 
1
0
x  x
N.B. lim
3x 2  2 x  7
x  5 x 2  x  1
(b) lim ( x 4  1  x 2 )
Example
Evaluate
(a) lim
Theorem
Let f (x) and g (x ) be two functions defined on R. Suppose X is a positive real number.
x 
(a) If f (x) is bounded for x > X and lim g ( x)  0 , then lim f ( x) g ( x)  0
x 
x 
(b) If f (x) is bounded for x > X and lim g ( x)   , then lim [ f ( x)  g ( x)]   ;
x 
x 
(c) If lim f ( x)  0 and f (x) is non-zero for x > X , and lim g ( x)   , then
x 
x 
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Continuity
Advanced Level Pure Mathematics
lim f ( x) g ( x)   .
x 
Evaluate (a)
Example
(d)
Theorem
sin x
x
xex
lim
x  1  x
(b) lim e x cos x
lim
x 
x 
(c) lim (e x  sin x)
x 
x  cos x
x 
x 1
(e) lim
SANDWICH THEOREM FOR FUNCTIONS
Let f(x) , g(x) , h(x) be three functions defined on R.
If lim f ( x)  lim h( x)  a and there exists a positive real number X such that when x > X ,
x 
x 
f ( x)  g ( x)  h( x) , then lim g ( x)  a .
x 
y
y = h(x)
y = f(x)
0
Example
x
(a) Show that for x > 0,
( x  1)( x  3)
 ( x  1)( x  3)  x  2 .
x2
(b) Hence find lim [ ( x  1)( x  3)  x] .
x 
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4
Continuity
N.B.
Advanced Level Pure Mathematics
1
lim (1  ) n  e for any positive integer n.
n 
n
Example
1
Evaluate lim (1  ) x  e by sandwich rule.
x 
x
Theorem
1
lim (1  ) x  e  lim (1  y) y  e
x 
y 0
x
1
Evaluate (a)
Example
N.B.
1.
lim (1  x)
 x2 1
(b) lim  2

x   x  1 


1
x
x0
1
lim (1  ) x  e
x  
x
x2
1
(c) lim x 1 x
x 1
1
2.
lim (1  x) x  e .
x 0
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5
Continuity
Advanced Level Pure Mathematics
Exercise
1
x
(a)
lim (1  3x)
x 0
2
(b) lim (1  )  x
x 
x
(c)
lim (1  tan x) cot x
(d) lim (
x
x 
x 1 x
)
x 1
B. Limit of a Function at a Point
Definition
Let f(x) be a function defined on R. lim f ( x)   means that for any  > 0 , there exists  > 0 such
xa
that when 0  x  a   , f (x)     .
y
y = f(x)
f(x)
l
a x
0
N.B.
x
(1) lim f ( x)   means that the difference between f(x) and A can be made arbitrarily small when x is
xa
sufficiently close to a.
(2) If f(x) is a polynomial , then lim f ( x)  f (a ) .
xa
(3) In general , lim f ( x)  f (a ) .
xa
(4) f(x) may not defined at x = a even through lim f ( x ) exists.
x a
lim cos x  cos   1 .
Example
lim ( x 2  2 x  5)  32  2(3)  5  8 ,
Example
3 if x  5
Let f (x)  
, then lim f ( x)  3 but f(5) = 1 ,
x 5
1
if
x
=
5

x 
x3
so lim f ( x)  f (5) .
x 5
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6
Continuity
Example
Advanced Level Pure Mathematics
2 x  1, x  2
Consider the function f ( x)  
.
x2
 4,
f (x ) is discontinuous at x  2 since lim f ( x)  lim (2 x  1)  5  f (2).
x 2
x 2
( x  2)( x  2)
x2  4
, then lim f ( x)  lim
 lim ( x  2)  4 .
x2
x 2
x 2
x2
x2
But f(2) is not defined on R.
Example
Let f ( x) 
N.B.
lim f ( x)  l is equivalent to :
xa
(i)
lim [ f ( x)  ]  0 ;
xa
(ii) lim f ( x)    0 ;
x a
(iii) lim f ( x)    lim f ( x);
xa
xa
(iv) lim f (a  h)   .
h 0
Theorem
Rules of Operations on Limits
Let f (x) and g (x ) be two funcitons defined on an interval containing a , possibly except a .
If lim f ( x)  h and lim g ( x)  k , then
xa
(a)
xa
lim  f ( x)  g ( x)  h  k
xa
(b) lim  f ( x) g ( x)  hk
xa
(c)
Example
lim
xa
f ( x) h
 , (if k  0)
g ( x) k
Evaluate (a)
(c)
x3  1
x 1 x  1
2 x
lim
x2
3  x2  5
lim
(1  x) n  1
x 0
x
(b) lim
(d)
lim ( x 2  x  x)
x  
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7
Continuity
Example
Example
Advanced Level Pure Mathematics
1 x
x 3 2  2 x 2
Evaluate (a)
lim
(c)
lim
x 2
x2
x2  2  2
1 x
x 1 2  2 x 2
(b) lim
(d) lim
(a) Prove that for x  ( 2 ,0)  (0, 2 ) ,
1 x
x 1
1 3 x
2
sin x tan x
2
.

1  cos x
cos x
sin x tan x
 2.
x 0 1  cos x
(b) Hence, deduce that lim
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8
Continuity
3.3
Advanced Level Pure Mathematics
Properties of Limit of a Functon
Theorem
Uniqueness of Limit Value
If lim f ( x)  h and lim f ( x)  k , then h  k .
xa
Theorem
xa
Heine Theorem
lim f ( x)  l  lim f ( x n )  l where x n  a as n   .
x a
n 
Example
Prove that lim sin
Example
If lim
Assignment
Ex.3A (1-6)
x 0
1
does not exist.
x
x 3  ax 2  x  4
exists, find the value of a and the limit value.
x  1
x 1
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9
Continuity
3.4
Advanced Level Pure Mathematics
Two Important Limits
sin x
1
x
Theorem
lim
N.B .
(1) lim
Example
Find the limits of the following functions:
Example
x 0
sin x
cos x
 lim

x 
x 
x
x
cos x
(3) lim

x 0
x
tan x
x 0
x
(a)
lim
(c)
lim ( x  ) sec x
x 2
2

x
tan x
 lim

x 0 sin x
x 0
x
1
(4) lim sin 
x 0
x
(2) lim
1  cos x
x 0
x2
(b) lim
(d) lim
x 0
sin 4 x
sin 5 x
Find the limits of the following functions:
(a)
lim
x 1
sin ( x  1)
x 2  3x  2
cos ax  cos bx
x 0
x2
(b) lim
(c)
3 sin x  sin 3x
x 0
x3
lim
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10
Continuity
Example
Advanced Level Pure Mathematics
Let  be a real number such that 0   
By using the formula sin

 2 sin

k 1
2
cos
 , show that
k 1
2
2
θ
θ
θ
θ
sin θ  2 n 1 cos cos 2 ... cos n 1 sin n-1 .
2
2
2
2
x
x
x
Hence, find the limit value lim (cos cos 2 ... cos n )
n 
2
2
2
Assignment
2
k
.
Ex. 3B (1-2)
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11
Continuity
3.5
Advanced Level Pure Mathematics
Left and Right Hand Limits
Theorem
lim f ( x)  l iff
lim f ( x)  lim f ( x)  l .
Example
Show that each of the following limits does not exist.
x a 
x a
(a)
lim
x 2
xa
x2
(b) lim
x 0
x
x
1
1
(c)
lim e x
x0
1
Example
By Sandwich rule, show that lim (a x  b x ) x does not exist for a  b  0 .
Solution
If a  b  0 then
x 0
b
1
a
1
If a  b  0 and x  0 then
1
1
b
If x  0 then ( ) x  1 x  1
a
1
x
1
x
1
b
 a  (a  b )  a{1  ( ) x } x  a  2 x
a
x
As lim 2  1 , by sandwich rule,
x
x0
1
x
1
we have lim (a x  b x ) x  b
we have lim (a  b ) x  a
x0
x 0
1
Since a  b, lim (a x  b x ) x does not exist.
x 0
Assignment
x0
1
1
x
1
1
b
a
( )x 1  ( )x 1
a
b
1
1
1
a x
x
x x
b  (a  b )  b{( )  1}x  b  2 x
b
x
As lim b  2  b , by sandwich rule,
( Why ? )
Ex. 3C (1-3)
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Continuity
Advanced Level Pure Mathematics
3.6
Continuous Functions
N.B.
In general, lim f ( x)  f (a ) .
xa
Definition
Let f(x) be a function defined on R. f(x) is said to be continuous at x = a if and only if
lim f ( x)  f (a ) .
xa
N.B.
This is equivalent to the definition :
A function f(x) is continuous at x = a , if and only if
(a) f(x) is well-defined at x = a , i.e. f(a) exists and f(a) is a finite value,
(b) lim f ( x ) exists and lim f ( x)  f (a ) .
xa
x a
Remark Sometimes, the second condition may be written as lim f (a  h)  f (a )
h 0
Example
1

Show that the function f ( x)   x sin x , x  0 is continuous at x = 0.
0
, x=0
A function is discontinuous at x = a iff it is not continuous at that point a.
y
y = f(x)
There are four kinds of discontinuity:
Definition
f(a)

(1) Removable discontinuity:
o
lim f ( x ) exists but not equal to f(a).
x a
0
Example
2 x  1, x  2
Show that f ( x)  
is discontinuous at x = 2.
, x=2
4
Solution
Since f(2) = 4 and lim f ( x)  lim (2 x  1)  5  f (2) ,
x2
a
x
x2
so f(x) is discontinuous at x = 2.
Example
1  cos x
Let f ( x)   x 2 , x  0 .
 a
, x=0
(a) Find lim f ( x ) .
x 0
(b) Find a if f(x) is continuous at x = 0.
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Continuity
y
(2) Jump discontinuity :

o
lim f ( x)  lim f ( x)
xa 
Advanced Level Pure Mathematics
xa
0
Example
Show that the function f(x) = [x] is discontinuous at 2.
Example
Find the points of discontinuity of the function g(x) = x  [x].
Example
 x 3  2 x  1, x  0
Let f ( x)  
.
sin
x
,
x

0

y = f(x)
a
x
Show that f(x) is discontinuous at x = 0.
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14
Continuity
Advanced Level Pure Mathematics
y
(3) Infinite discontinuity:
lim f ( x)   or lim f ( x)  
x a
y = f(x)
xa
i.e. limit does not exist.
0
Show that the function f ( x) 
Example
a
x
1
is discontinuous at x = 0.
x
y
f ( x)  e
(4) At least one of the one-side limit does not exist.
1
x
lim f ( x) or lim f ( x) do not exist.
xa 
xa
1
1
Since lim e x  0 and lim e x   ,
Example
x0
x
x0
1
so the function f ( x)  e x is discontinuous at x = 0.
(1) A function f(x) is called a continuous function in an open interval (a, b) if it is continuous at
every point in (a, b).
Definition
(2) A function f(x) is called a continuous function in an closed interval [a, b] if it is continuous at
every point in (a, b) and lim f ( x)  f (a ) and lim f ( x)  f (b) .
x b
xa
(1)
(2)
y
y
y

o
o
a
N.B.
b
x
a
b

a
x
b
f(x) is continuous
f(x) is continuous
f(x) is continuous
on (a, b).
on (a, b).
on [a, b].
(1)
f (x ) is continuous at x = a  lim f ( x)  f (a)  lim f (a  h)  f (a) .
(2)
f (x ) is continuous at every x on R  lim f ( x  h)  f ( x) for all x  R.
xa
x
h 0
h 0
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Continuity
Example
Advanced Level Pure Mathematics
Let f (x) be a real-value function such that
(1)
f ( x  y )  f ( x) f ( y ) for all real numbers x and y ,
(2)
f (x ) is continuous at x  0 and f (0)  1 .
Show that f (x) is continuous at every x  R.
3.7
A.
Theorem
Properties of Continuous Functions
Continuity of Elementary Functions
Rules of Operations on Continuous Functions
If f(x) and g(x) are two functions continuous at x = a , then so are f ( x)  g ( x) , f ( x) g ( x)
and
f ( x)
provided g(a)  0.
g ( x)
1  cos x
. Since f ( x)  1  cos x and g ( x)  x 2 are two functions continuous
2
x
everywhere, but g(0) = 0, so h is continuous everywhere except x = 0.
Example
Let h( x) 
Example
Let h( x) 
Theorem
Let g(x) be continuous at x = a and f(x) be continuous at x = g(a), then f。g is continuous at x = a.
Example
sin e x , e sin x are continuous functions.
Example
Show that f ( x)  cos x 2 is continuous at every x  R.
1  cos x
. Since f ( x)  1  cos x and g ( x)  x 2  1 ( 0) are two functions
x2 1
continuous everywhere, so h is continuous everywhere.
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Continuity
Example
Advanced Level Pure Mathematics
Let f and g be two functions defined as f ( x) 
 x if x  0
x | x |
for all x and g ( x)   2
.
2
 x if x  0
For what values of x is f [g(x)] continuous?
Theorem
Let lim f ( x)  A and g(x) be an elementary function (such as sin x, cos x, sin 1 x, e x , ln x, ...
x 
etc.) for which g(A) is defined, then
lim g[ f ( x)]  g[lim f ( x)]  g ( A)
x 
Example
lim cos( tan x)  cos(lim tan x) 
Example
Evaluate lim
Example
Evaluate lim tan (
x 1
x 
x 1
ln( 1  x)
.
x 
x
x 0
sin x
).
x
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Continuity
Advanced Level Pure Mathematics
B. Properties of Continuous Functions
(P1)
If f(x) is continuous on [a, b], then f(x) is bounded on [a, b].
(P2)
But f(x) is continuous on (a, b) cannot imples that f(x) is bounded on (a, b).
(1)
y
c < f(x) < d
y
f(x) is not
bounded
on (a, b).

d
c

0
a
0
b
a
b
x
x
f(x) = cot x is continuous on (0, ) , but it is not bounded on (0, ).
Example
(P3)
(2)
If f(x) is continuous on [a, b], then it will attain an absolute maximum and absolute minimum
on [a, b].
i.e. If f(x) is continuous on [a, b], there exist x 1  [a, b ] such that f ( x 1 )  f ( x )
and x 2  [a, b ] such
that f ( x 2 )  f ( x ) for all x  [a , b ] . x 1 is called the absolute maximum of the function and x2 is
called the absolute minimum of the function.
no. abs. max. on [a, b]
y
o
abs. max. on [c, d]
|
o

0

a
b
c
d



abs. min. on [c, d]

abs. min. on [a, b]
(P4)
If f(x) is continuous on [a, b] and f(a)f(b) < 0 , then there exists c  [a, b] such that f(c) = 0.
y
y

0
a

b
x
0
a
b

x

Prepared by K. F. Ngai
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Continuity
Advanced Level Pure Mathematics
Example
Let f ( x)  cos x  0.6 which is continuous on [0, 2 ] and f(0) = 0.4 and f ( 2 )  0.6 . Hence,
there exists a real number c  [0, 2 ] such that f(c) = 0.
Example
Prove that the equation x  e x  2 has at least one real root in [0, 1].
Example
Let f ( x)  x 3  2 x .

(a) Show that f(x) is a strictly increasing function.
(b) Hence, show that the equation f ( x)  0 has a unique real root.
Prepared by K. F. Ngai
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Continuity
(P5)
Advanced Level Pure Mathematics
Intermediate Value Theorem
If f(x) is continuous on [a, b], then for any real number m lying between f(a) and f(b) , there
corresponds a number c  [a, b] such that f(c) = m.
y

f(b)
m
f(a)
0
(P6)
Example

a c
b
x
Let f(a) = c and f(b) = d, if f is continuous and strictly increasing on [a, b], then f
continuous and strictly increasing on [c, d] ( or [d, c] ).
1
is also
f ( x)  x 2 is strictly increasing and continuous on [0, 5], so f 1 ( x)  x is also strictly
increasing and continuous on [0, 25].
Example
f ( x)  cos x is strictly decreasing and continuous on [0,π], so f 1 ( x)  cos 1 x is also
strictly decreasing and continuous on [1, 1].
Example
(a) Suppose that the function satisfies f(x+y) = f(x) + f(y) for all real x and y and f(x) is
continuous at x = 0. Show that f(x) is continuous at all x.
(b) A function
 1  sin x  1  sin x

if x  0 .
f ( x)  
x

if x = 0
 0
Is f(x) continuous at x = 0? If not , how can we redefine the value of f(x) at x = 0 so that it is
continuous at x = 0?
Prepared by K. F. Ngai
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Continuity
Advanced Level Pure Mathematics
Exercise
Example
Let f(x) be a continuous function defined for x > 0 and for any x , y > 0,
f(xy) = f(x) + f(y).
(a) Find f(1).
(b) Let a be a positive real number. Prove that f (a r )  r  f (a) for any non-negative rational
number.
(c) It is known that for all real numbers x, there exists a sequence {xn } of rational numbers such
that lim x n  x .
x 
(i) Show that f (a x )  x  f (a) for all x > 0, where a is a positive real constant.
(ii) Hence show that f ( x)  c ln x for all x > 0, where c is a constant.
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Continuity
Example
Advanced Level Pure Mathematics
Let f be a real-valued continuous function defined on the set R such that
f(x+y) = f(x) + f(y) for all x , y  R.
(a) Show that
(i) f(0) = 0
(ii) f(x)= f(x) for any x  R.
(b) Prove that f(nx) = nf(x) for all integers n.
Hence show that f(r) = rf(1) for any rational number r.
(c) It is known that for all x  R, there exists a sequence {a n } of rational numbers such that
lim a n  x .
n 
Using (b), prove that there exists a constant k such that f(x) = kx for all x  R.
Example
Let R denote the set of all real numbers and f : R  R a continuous function not identically zero
such that f(x+y) = f(x)f(y) for all x , y  R.
(a) Show that
(i) f(x)  0 for any x  R.
(ii) f(x) > 0 for any x  R.
(iii) f(0) = 1
(iv) f ( x)   f ( x)1
(b) Prove that for any rational number r, f (rx )   f ( x)r .
Hence prove that there exists a constant a such that f ( x)  a x for all x  R.
Assignment
Ex.3D (1-9).
Revision Exercise (1-6).
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