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Continuity
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
3
Calculus I
Chapter 3
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(2003)
Page
1
Continuity
Advanced Level Pure Mathematics
3.1
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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Continuity
Advanced Level Pure Mathematics
Theorem
UNIQUENESS of Limit Value
If lim f ( x)  a and lim f ( x)  b , then a  b .
x 
Theorem
x 
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 
N.B. lim
x 
x 
n
f ( x)  n lim f ( x)
x 
1
0
x
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 
lim f ( x) g ( x)   .
x 
Prepared by K. F. Ngai(2003)
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Continuity
Advanced Level Pure Mathematics
Example
Evaluate (a)
lim
sin x
x 
x
(b) lim e x cos x
Example
Evaluate (a)
lim
xex
x  1  x
(b) lim
Theorem
SANDWICH THEOREM FOR FUNCTIONS
x 
(c) lim (e x  sin x)
x 
x  cos x
x 
x 1
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
x
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Continuity
Advanced Level Pure Mathematics
Example
(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.
Example
1
lim (1  ) n  e for any positive integer n.
n 
n
1
Evaluate lim (1  ) x  e by sandwich rule.
x 
x
Prepared by K. F. Ngai(2003)
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Continuity
Advanced Level Pure Mathematics
1
1
lim (1  ) x  e  lim (1  y) y  e
x 
y 0
x
Theorem
Evaluate (a)
Example
N.B.
Exercise
1.
lim (1  x)
x0
1
lim (1  ) x  e
x  
x
(a)
lim (1  3x)
x 0
 x2 1
(b) lim  2

x   x  1 


1
x
1
x
x2
1
(c) lim x 1 x
x 1
1
2.
lim (1  x) x  e .
x 0
2
(b) lim (1  )  x
x 
x
(c)
lim (1  tan x) cot x
x 
(d) lim (
x 
x 1 x
)
x 1
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Continuity
Advanced Level Pure Mathematics
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) . Therefore, f is discontinuous at x  5 .
x 5
Example
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
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Continuity
Advanced Level Pure Mathematics
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
x a
(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)
Example
x3  1
x 1 x  1
2 x
lim
x2
3  x2  5
lim
1 x
x 3 2  2 x 2
Evaluate (a)
lim
(c)
lim
x 2
x2
x2  2  2
(1  x) n  1
x 0
x
(b) lim
(d)
lim ( x 2  x  x)
x  
1 x
x 1 2  2 x 2
(b) lim
(d) lim
x 1
1 x
1 3 x
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Continuity
Advanced Level Pure Mathematics
Example
(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
Prepared by K. F. Ngai(2003)
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9
Continuity
Advanced Level Pure Mathematics
3.3
Theorem
Properties of Limit of a Functon
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
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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Continuity
Advanced Level Pure Mathematics
Exercise 3A
1. Give an example to show that the existence of lim [ f ( x)  g ( x)] doex not imply the existenceof
xa
lim f ( x ) or lim g ( x) .
x a
2.
x a
Give an example to show that the existence of lim f ( x) g ( x) doex not imply the existence of lim f ( x ) or
x a
x a
lim g ( x) .
x a
3.
Evaluate the following limits.
x  cos x
3x
2x 
(a) lim
(b) lim 


x 
x


x 1
 x  1 x  1
(e)
lim
x 1
2 x 2
x2
(f)
2x3
x  x 2  1
lim
(c)
x 2  3x  2
lim ( x 4  1  x 2 ) (d) lim 2
x 1 x  5 x  6
x
(g)
lim ( x 2  1  x) (h) lim
x
(1  x) 4  (1  4 x)
x 0
x  x4
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Continuity
Advanced Level Pure Mathematics
3.4
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
3 sin x  sin 3x
x 0
x3
(Hint: sin 3 A  3 sin A  4 sin 3 A )
(c)
lim
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Continuity
Advanced Level Pure Mathematics
Example
Let  be a real number such that 0   
By using the formula sin

2
k
 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
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Continuity
Advanced Level Pure Mathematics
Exercise
1. Evaluate each of the following limits, if any.
sin 5 x
sin 6 x
(a) lim
(b) lim
x 0
x  0 sin 2 x
3x
2
sin x
sin x
(c) lim
(d) lim
x    x
x 0
x
2.
Evaluate each of the following limits, if any.
(a)
lim (1  tan x)
(c)
 x 2  2x 

lim  2
x 
x

1


cot x
x 
 x 2  2x  3 
(b) lim  2

x   x  3 x  28 


x
x
(d) lim (cos x) cot
2
x
x 0
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Continuity
Advanced Level Pure Mathematics
3.5
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
x
x0
1
1
we have lim (a x  b x ) x  b
we have lim (a  b ) x  a
x0
x0
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,
1
Since a  b, lim (a x  b x ) x does not exist.
x 0
( Why ? )
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Continuity
Advanced Level Pure Mathematics
Exercise
1. Determine whether the following limits. If so, find its limit.
x2
x 
1
(a) lim
(b) lim
(c) lim
1
x0 x
x2 x  2
x 0
3 4x
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
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Continuity
Advanced Level Pure Mathematics
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.
4
,
x
=
2

Solution
Since f(2) = 4 and lim f ( x)  lim (2 x  1)  5  f (2) ,
x2
x
a
x 2
so f(x) is discontinuous at x = 2.
1  cos x
Let f ( x)   x 2 , x  0 .
 a
, x=0
Example
(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
Example
a
y = f(x)
x
Show that the function f(x) = [x] is discontinuous at 2.
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Continuity
Advanced Level Pure Mathematics
Example
Find the points of discontinuity of the function g(x) = x  [x].
Example
 x 3  2 x  1, x  0
Let f ( x)  
.
x0
sin x,
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
Example
Show that the function f ( x) 
a
x
1
is discontinuous at x = 0.
x
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Continuity
Advanced Level Pure Mathematics
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a
x b
(1)
(2)
y
y
y

o
o
a
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].
N.B.
Example
(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
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.
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Continuity
Advanced Level Pure Mathematics
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.
Example
Let f and g be two functions defined as f ( x) 
1  cos x
. Since f ( x)  1  cos x and g ( x)  x 2  1 ( 0) are two functions
2
x 1
continuous everywhere, so h is continuous everywhere.
 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 
x 
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Continuity
Advanced Level Pure Mathematics
Example
lim cos( tan x)  cos(lim tan x) 
Example
Evaluate lim
Example
Evaluate lim tan (
x 1
x 1
ln( 1  x )
.
x 0
x
x 0
sin x
).
x
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
Example
(2)
c

0
a
0
b
a
b
x
x
f(x) = cot x is continuous on (0, ) , but it is not bounded on (0, ).
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Continuity
Advanced Level Pure Mathematics
If f(x) is continuous on [a, b], then it will attain an absolute maximum and absolute minimum
on [a, b].
(P3)
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
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].

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Continuity
Advanced Level Pure Mathematics
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.
(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

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].
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Continuity
Advanced Level Pure Mathematics
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?
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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 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.
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Continuity
Advanced Level Pure Mathematics
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.
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Continuity
Advanced Level Pure Mathematics
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.
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Continuity
Advanced Level Pure Mathematics
Example
(a) Let x  1 and define a sequence a n  by
a n 1  1
for n  2
2a n 1
(i) Show that a n  1 and an  an1 for all n .
(ii) Show that lim a n  1 .
2
a1  x
and a n 
n 
(b) Let f : [1, )  R be a continuous function satisfying
x2 1
for all x  1.
)
2x
Using (a), show that f ( x)  f (1) for all x  1.
f ( x)  f (
(1996 Paper II)
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