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CHAPTER 02
LIMITS AND CONTINUITY
1
...
Hence we must develop some understanding and insight
of this concept
...
In lay term we can say a limit means a
boundary: a point, line or number that may not be passed or crossed
...

Example 1
Consider the perimeter (circumference) of a circle
...
The perimeter of the triangle is less than the
perimeter of the circle
...
The perimeter of the square is less than that of
the circle
...
The perimeter will be less than that of the circle
...
However, it will never be equal to that of the circle
...

Example 2
Suppose we consider the function f defined by
2x 2  5x  2
f(x) =
x2

14

Observe that f(x) is defined for all values of x except when x = 2, for at that point the
denominator is zero
...

Let us now closely examine the behavior of f(x) for values of x “near 2”
...
7) = 2
...
8) = 2
...
9) = 2
...
99) = 2
...
999) = 2
...
3) = 3
...
2) = 3
...
1) = 3
...
01) = 3
...
001) = 3
...
To express
this situation, we say that 3 is the limit of f as x approaches 2
...

x 2
x2

lim

Actually, by taking any value of x sufficiently close to 2, we can make the number f(x) as
arbitrarily close to 3 as we wish
...

Since 2x2 – 5x = 2 = (2x-1)(x-2), then
f(x) 

(2x  1)(x  2)

...
7) = 2
...
8) = 2
...
9) = 2
...
99) = 2
...
999) = 2
...
3) = 3
...
2) = 3
...
1) = 3
...
01) = 3
...
001) = 3
...
Thus, we see that if x is sufficiently close to 2, the number (2x – 1) will be
as close to 3 as we please
...


15

y

Fig B

f(x) = 2x-1, x≠2
3

x

0

2

Example 3
Let us determine the limit of the function f(x) = x2 as x approaches -1
...


y

Fig C

f(x) = x2

1

x

0

-1

Thus,

lim x 2  1

x 1

Remark: Note that in determining this limit, we are not at all concerned with what
happens to the function when x equals -1
...

Again, we are not concerned with what happens to f(x) when x = a, but only with what
happens to it when x is close to a
...
We emphasize that a limit is
x2
independent of the manner in which x approaches a
...
e
...


Here follow some theorems of limits
...
We
shall prove them later in the next chapter
...


If f(x) = c is a constant function , then lim f(x)  lim c  c

2
...


x a

lim[f(x)  g(x)]
x a

=

lim f(x) ± lim g(x)

=

L1 ± L2

xa

xa

This property can be extended to the limit of any finite number of sums or
differences
...


5
...
g(x) ]
xa

lim[c
...
lim g(x)

=

L1
...
lim g(x)

=

c
...


xa

xa

xa

17

6
...


=

lim n f(x)
x a

lim f(x)

n

x a

=

n

L1

We can summarize these theorems
1
...

2
...

3
...

4
...

5
...






6
...

x a

n

x a

7
...

x a

8
...

x a

9
...

x a

10
...
( if n is even, we assume a  0 )
...
lim n f ( x)  n lim f ( x) where n is a positive integer
...

x a

Examples
(a)

lim (x 3  x 2 ) =
x 3

lim x 3
x 3

=

33

=

36

 lim x 2
x 3

+ 32

18

(b)

(c )

(d)

(e)

lim (x 2  4x  2)
x 4

lim (x 2  1)(x 3  4)
x 1

lim [3 (x  4)]

x  2

t3  t2  2
lim
t 0
t3 1

=

lim x 2  lim 4x  lim (2)

=

4

2

=

30

x 4

x 4

+ 4(4) - 2

lim (x 2  1) lim (x 3  4)

=

lim x

=

[1 + 1][1- 4]

=

-6

=

3 lim (x  4)

=

3(-2 -4)

=

-18

x 1

x 1

2

x 1





 lim 1 lim x 3  lim 4
x 1

x 1

x 1

x  2

lim (t 3  t 2  2)
=

t 0

lim (t 3  1)
t 0

=

lim x 2  5

x 4

=

=

(f)

x 4

=

2
1
2
lim (x 2  5)
x 4

=

lim x 2  lim 5

=

16  5

=

21

x 4

x 4

In all the examples above, we could evaluate the limits by direct substitution
...
Here follow some of those “tricks”:

19

(g)

(h)

x2  9
x 3 x 2  4x  3

(x  3)(x  3)
x 3 (x  3)(x  1)

= lim

lim

lim
x 0

(x  3)
x 3 (x  1)

=

lim

=

6
2

=

3

4x 2
x

In this case factorization will not be possible
...
This does not mean that ∞ (read “infinity”) is the
0

limit, but that the limit does not exist because the quotient becomes large in size
as x approaches 5
...

(iii)

x7
= -∞
x  2 (x  2) 2

lim

Since the quotient becomes large negative as x approaches 2, the expression
becomes negatively infinite
...

1
...


lim (x 3  4x  1)

x  1

lim

t  1

1  2t
3t  21

x 3  3x 2  10x
x 2
x2

5
...


x 2  5  30
lim
x 5
x 5
1



1
2

9
...


lim

1 t  1 t
t

x 2

t 0

x2 1
x 1 x  1

2
...


lim

6
...


10
...


12
...
(a) Estimate the value of

lim
x 0

x
1  3x  1

By graphing the function f (x) 

x
1  3x  1


...

(c) Use the limit laws to prove that your guess is correct
...


ONE-SIDED LIMITS

Sometimes the values of a function f(x) tend to different limits as x approaches a number
c from different sides
...
Similarly we call the limit of f(x) as x
approaches c from the left, the left-hand limit of f(x) at c
...

Definition “Informal”:
Let L be a real number
(i)
Suppose that f(x) is defined near c for x > c, and that as x gets close to c, f(x)
gets close to L
...

Then we say that L is the left-hand limit of f(x) as x
approaches c from the left, and we write
lim f(x)  L
x c

Theorem 2
...
02 + 4

=

4

=

lim (t  2) 2

t 0 

=

(0-2)2

=

(-2)2

=

4

Therefore , lim g(t)  4 since lim g(t) = lim g(t) = 4
...


k  1

lim (k  7)

k  3-

=

-3 + 7

=

4

=

lim k  2

k   3

=

│-3 - 2│

=

│-5│

=

5

Therefore lim h(k) does not exist, since lim - h(k) ≠ lim  h(k)
k  3

k  3

k  3

EXERCISES

1
...


3
...


x 1

  2t  4

f (t )   4
 t 2 1


Let

Find

lim f(x)

x  1

lim g(x)
x 1

(ii)

;

x  1

; 1  x  1
;

1 x  2

;

x2

lim g(x)
x 2

CONTINUITY

In ordinary language to say a certain process is “continuous” is to say that the process
goes on without interruption and without abrupt changes or without a break
...
Intuitively speaking, a
function is continuous at a point if its graph does not have a break or jump at that point
...

y
3
f(x) = 2
1
x

0

We cannot always rely on the drawing of graphs for continuity of functions, because
sometimes it is difficult, if not impossible, to draw some of the graphs
...

Definition 2
...

(ii)
lim f(x) exists
xc

lim f(x) = f (c )

(iii)

xc

Remark:
The limit in the definition above is to be a two-sided limit
...


1
...


25

2
...


Solution:
(i)
(ii)

g(2) = 4
Since g is defined in the same way on the left-hand side of 2 as on
the right hand side of 2, we need only find the two-sided limit
x2  4
lim g(x) = lim
x2
x 2 x  2
(x  2)(x  2)
= lim
x 2
x2
= lim (x  2)
x 2

=
=

2+2
4
...

x2

(iii)

Since lim g(x) = g(2) = 4, g is continuous at x = 2
...
3:
A function is continuous over (or in) the open interval (a,b) if f is continuous at every
point in that interval
...
4:
The function f is continuous in the closed interval [a,b] if all the following conditions
hold:
(i)
f is continuous at every x in the interval (a,b)
(ii)
f(a) and f(b) both exist
...

x b

26

Example:
Determine whether the function h defined by
x  1
 1 ;
 x3
; 1  x  1

h(x)  
 1 x ; 1  x  2
3  x 2 ;
x2
is continuous at (a) x = -1
(b) the interval [1,2]
Solution:
(a)
(i)
h(-1) = (-1)3 = -1
(ii)
lim  h(x) = lim  (1) = -1
x  1

x  1

lim h(x) =

x  1

(iii)
(b)

lim x 3 = (-1)3 = -1

x  1

lim h(x) = h(-1) = -1

x  1

Hence h is continuous at x = -1
...
Hence
lim h(x) = lim (1  x) = 1- c
x c

x c

and therefore h(c ) = 1-c = lim h(x) , which gives h continuous at every
x c

(ii)
(iii)

point c  (1,2)
...

lim h(x) = lim (1  x) = 1-1 = 0
x 1

x 1

Hence lim h(x) ≠ h(1) and h is not right continuous at x = 1
...

x 2

Conclusion: h is continuous over the interval (1,2]
...
5:
If the function f is not continuous at x = c, then f is said to be discontinuous at c
...
2) of continuity fails to hold
...


27

The discontinuity as a consequence of (i) is called an essential discontinuity
The discontinuity as a consequence of (ii) is called a removable discontinuity
...
If the limit is L, we
define f(c ) to be L
...
Can we make it continuous at x = 2?
Solution:
For f to be continuous at x = 2, lim f(x) = f(2)
...
e
...

2

lim f(x) = lim

The extended function
f(x) ; x  2


g(x)   5
; x2

 4

x2  x  6
; x2
 2
x

4
= 
5

; x2
4

The function g is called the continuous extension of the function f to the point x = 2
...
6
If the functions f and g are continuous at x = c, then
(i)
f + g is continuous at x = c
(ii)
f – g is continuous at x = c
(iii) f
...
g is continuous at x = c, (k any real number)

28

(v)

f
is continuous at x = c, if g(c ) ≠ 0
...
g)(x)
xc

= lim[f(x)
...
lim g(x)
xc

xc

= f (c)
...
g)(c)
(iv)

lim (k
...
g(x)]
xc

=

k
...
g(c)
= (k
...

=
g
Similar arguments with right-hand and left-hand limits establish the theorem for
continuity at end-points
...

1 ; x0

1
...

f(x)   0 ; x  0
 1 ; x  0


2
...


 x2
; x  0


g(x) =  2
; x  0
 x  1 ; x  0



at x = 0

4
...

6
...

a)

b)

c)
7
...


30

Continuous functions have two properties which are particularly important in calculus:
the Maximum-Minimum Value property and the Intermediate Value property
...
No proofs will be given, as proofs depend on the precise definition of a
limit which will be discussed in the next chapter
...
6:
A set of numbers S is bounded above if there exists a number M such that x ≤ M for all
x  S
...
The least upper bound of S is the
smallest upper bound
...
Such an m is called a lower bound
...


Example:
Consider the set S = { x | 0 < x <1} = (0,1)
M = 2 is an upper bound, but M = 1 is the least upper bound
...

Theorem 2
...
Then f takes on
a minimum value m and a maximum value M on [a,b]
...


y
M

m
0

x=a

x

x=b

31

x=a

y

x=b

M

x
0
m

REMARKS:
1
...

1
For example, f(x) =
is continuous on (0,1), but as x approaches 0 from the
x
right, f(x) becomes very large and f is therefore not bounded on that interval
...

2
...

1
For example, f(x) =
is not bounded on the interval [-1,1] because it is not
x
continuous at x = 0
...
8 (Intermediate Value Theorem)
Let f be a continuous function on [a,b]
...


32

y

y = f(x)

f(b)

k
f(a)

x
0

a

c

b

EXERCISES
1
...

(i)

f(x) = x2 -5 on [2,3] ; k = 2

(ii) f(x) = x2 + x + 1 on [ -2,3] ; k = 6
(iii) f(x) = x3 -2x + 1
(iv) f(x) =

2
...


3
...

Use the Intermediate Value Theorem to argue that Mpho was at some time in her
life exactly 122 centimeters tall

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