Download Continuity

Continuity
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
8
Calculus I
Chapter 8
Continuity
8.1
Introduction
2
8.2
Limit of a Function
2
8.3
Properties of Limit of a Function
7
8.4
Two Important Limits
8
8.5
Left and Right Hand Limits
9
8.6
Continuous Functions
13
8.7
Properties of Continuous Functions
14
Page
1
Continuity
Advanced Level Pure Mathematics
8.1
Introduction
Example f ( x )  x 2 is continuous on R.
Example
y
f ( x) 
y
y  x2
0
8.2
1
is discontinuous at x = 0.
x
y
x
0
1
x
x
Limit of a Function
A. Limit of a Function at Infinity
Defintion 1
Let f(x) be a function defined on R. lim f ( x)   means that for any  > 0 , there exists X > 0
x 
such that when x > X ,
f (x)     .
y
y
l+
l
l
l+
l
0
N.B. (1)
(2)
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.
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
Theorem 2
x
0
x
UNIQUENESS of Limit Value
If lim f ( x)  a and lim f ( x)  b , then a  b .
x 
x 
Page
2
x
Continuity
Advanced Level Pure Mathematics
Theorem 3
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 
f ( x)
f ( x) lim
 x 
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 1
Evaluate
(a) lim
Theorem 4
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 
lim f ( x) g ( x)   .
x 
Example 2
Evaluate (a)
sin x
x 
x
lim
(b) lim (e x  sin x)
x 
Page
3
x  cos x
x 
x 1
(c) lim
Continuity
Advanced Level Pure Mathematics
Theorem 5
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 3
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 
N.B.
1
lim (1  ) n  e for any positive integer n.
n 
n
Example 5
1
Evaluate lim (1  ) x  e by sandwich rule.
x 
x
Page
4
Continuity
Advanced Level Pure Mathematics
1
Theorem 6
1
lim (1  ) x  e  lim (1  y) y  e
x 
y 0
x
lim (1  x)
Example 6
Evaluate (a)
N.B.
1
lim (1  ) x  e
x  
x
1.
x 0
1
Exercise 7
 x2 1
(b) lim  2

x   x  1 


1
x
2.
x 1
lim (1  x)  e .
x 0
lim (1  3x) x
x 0
2
(b) lim (1  )  x
x 
x
(c)
lim (1  tan x) cot x
(d) lim (
x 
1
(c) lim x 1 x
1
x
(a)
x 
x2
x 1 x
)
x 1
B. Limit of a Function at a Point
Definition 7 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
Page
5
Continuity
Advanced Level Pure Mathematics
lim cos x  cos   1 .
lim ( x 2  2 x  5)  32  2(3)  5  8 ,
Example
8
Example 9
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
2 x  1, x  2
Example 10 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 11 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 8
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
x a
(a)
lim  f ( x)  g ( x)  h  k
x a
(b) lim  f ( x) g ( x)  hk
xa
(c)
lim
x a
f ( x) h
 , (if k  0)
g ( x) k
Example 12 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)
Page
6
lim ( x 2  x  x)
x  
Continuity
Advanced Level Pure Mathematics
Example 13 (a) Prove that for x  ( 2 ,0)  (0, 2 ) ,
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
8.3
Properties of Limit of a Functon
Theorem 9
Uniqueness of Limit Value
If lim f ( x)  h and lim f ( x)  k , then h  k .
x a
x a
Theorem 10 Heine Theorem
lim f ( x)  l  lim f ( x n )  l where x n  a as n   .
xa
n 
Example 14 Prove that lim sin
x 0
1
does not exist.
x
Page
7
Continuity
Advanced Level Pure Mathematics
x 3  ax 2  x  4
exists, find the value of a and the limit value.
x  1
x 1
Example 15 If lim
Assignment
8.4
Ex.8.1A (1-6)
Two Important Limits
sin x
1
x 0
x
Theorem 11
lim
N.B .
(1) lim
sin x
cos x
 lim

x 
x 
x
x
cos x
(3) lim

x 0
x
x
tan x
 lim
1
x 0 sin x
x 0
x
1
(4) lim sin 
x 0
x
(2) lim
Find the limits of the following functions:
Example 16
tan x
x
(a)
lim
(c)
lim ( x  ) sec x
x 2
2
x 0

(b) lim
1  cos x
x2
(d) lim
sin 4 x
sin 5 x
x 0
x 0
Page
8
Continuity
Advanced Level Pure Mathematics
Example 17 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
8.5
2
k
.
Left and Right Hand Limits
Theorem 12
lim f ( x)  l iff
Example
Show that each of the following limits does not exist.
x a
(a)
lim
x 2
x2
lim f ( x)  lim f ( x)  l .
x a 
x a
(b) lim
x 0
x
(c)
x
Page
9
1
lim e x
x0
Continuity
Advanced Level Pure Mathematics
1
1
Example 18 By Sandwich rule, show that lim (a x  b x ) x does not exist for a  b  0 .
x 0
Solution
If a  b  0 then
b
1
a
1
If a  b  0 and x  0 then
1
1
1
1
1
b
a  (a x  b x ) x  a{1  ( ) x }x  a  2 x
a
As lim 2 x  1 , by sandwich rule,


x0
1
x
1
x
x0
1
x0
x0
1
Since a  b, lim (a x  b x ) x does not exist.
( Why ? )
x 0
8.6
Continuous Functions
N.B.
In general, lim f ( x)  f (a ) .
1
we have lim (a x  b x ) x  b
we have lim (a  b )  a
x
1
1
b
a
( )x 1  ( )x 1
a
b
1
1
1
a
b  (a x  b x ) x  b{( ) x  1}x  b  2 x
b
As lim b  2 x  b , by sandwich rule,

b
If x  0 then ( ) x  1 x  1
a
xa
Definition 13 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
1

Example 19 Show that the function f ( x)   x sin x , x  0 is continuous at x = 0.
0
, x=0
Definition 14 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:
f(a)

(1) Removable discontinuity:
o
lim f ( x ) exists but not equal to f(a).
x a
0
2 x  1, x  2
is discontinuous at x = 2.
Example 20 Show that f ( x)  
4
,
x
=
2

Solution
Since f(2) = 4 and lim f ( x)  lim (2 x  1)  5  f (2) ,
x2
x 2
Page
10
a
x
Continuity
Advanced Level Pure Mathematics
so f(x) is discontinuous at x = 2.
1  cos x
Example 21 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.
y
(2) Jump discontinuity :

o
lim f ( x)  lim f ( x)
xa 
xa
0
y = f(x)
a
x
Example 22 Show that the function f(x) = [x] is discontinuous at 2.
 x 3  2 x  1, x  0
.
Example 23 Let f ( x)  
sin
x
,
x

0

Show that f(x) is discontinuous at x = 0.
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
Page
11
a
x
Continuity
Advanced Level Pure Mathematics
Example 24 Show that the function f ( 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
Example
25
1
Since lim e x  0 and lim e x   ,
x0
x
x0
1
so the function f ( x)  e x is discontinuous at x = 0.
Definition 15 (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).
(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a
x b
(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
h 0
h 0
Example 26 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.
Page
12
x
Continuity
Advanced Level Pure Mathematics
8.7
A.
Properties of Continuous Functions
Continuity of Elementary Functions
Theorem 16 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
x2
everywhere, but g(0) = 0, so h is continuous everywhere except x = 0.
Example 26 Let h( x) 
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.
Example 27 Let h( x) 
Theorem 17 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.
Theorem 18 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 28
x 
lim cos( tan x)  cos(lim tan x) 
x 1
x 1
ln( 1  x)
.
x 
x
Example 29 Evaluate lim
Example 30 Evaluate lim tan (
x 0
sin x
).
x
Page
13
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

Example 31 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 32 Let f ( x)  x 3  2 x .
Page
14
Continuity
Advanced Level Pure Mathematics
(a) Show that f(x) is a strictly increasing function.
(b) Hence, show that the equation f ( x)  0 has a unique real root.
(P5)
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 33

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 34
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].
Exercise
Example
Let f(x) be a continuous function defined for x > 0 and for any x , y > 0,
Page
15
Continuity
Advanced Level Pure Mathematics
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 xn  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.
Example
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 {an } 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.
Page
16