1
LIMIT OF A FUNCTION
I.
Intuitive Notion of the Limit of a Function
1. Consider the linear function
f ( x)  x  3 whose graph is shown in Figure 1.
y-axis
-3
x-axis
-6
x
f (x)
-3.5
-6.5
-3.2
-6.2
-3.1
-6.1
Figure 1. The graph of f.
-3.01
-6.01
-3.001
-6.001
-2.999
-5.999
-2.99
-5.99
-2.9
-5.9
-2.8
-5.8
-2.5
-5.5
Observe that as x approaches  3 , either from the left or from the right, f (x) approaches
 6 . This means the limit of f (x) as x approaches  3 is equal to  6 . We write this in
symbol as lim f ( x )  6 .
x  3
2. Consider the rational function g ( x) 
x2  9
whose graph is shown in Figure 2.
x3
y-axis
-3
x-axis
-6
Figure 2. The graph of g.
2
x
g (x)
-3.5
-6.5
-3.2
-6.2
-3.1
-6.1
-3.01
-6.01
-3.001
-6.001
-2.999
-5.999
-2.99
-5.99
-2.9
-5.9
-2.8
-5.8
-2.5
-5.5
Observe that as x approaches  3 , either from the left or from the right, g (x) approaches
g ( x)  6 . The
 6 . This means the limit of g (x) as x approaches  3 is equal to  6 , or xlim
 3
only difference between this case and the preceding example is that g is not defined at x  3
as indicated by the gap in the graph; elsewhere functions f and g are equal.
Remark: In finding the limit of a function F as x approaches a value c , we only need to
consider x values sufficiently close to c and observe how F behaves at these x values. It
is not necessary that F is defined at x  c inorder for the limit of F to exist there. The limit of
F exists as long as it approaches the same value as x approaches c from either from the right or
from the left.
 x2  9

if x  3
3. Consider the piecewise-defined function h( x)   x  3
whose graph is shown


2
if
x


3

in Figure 3.
y-axis
-3
x-axis
-6
x
h(x)
-3.5
-6.5
-3.2
-6.2
-3.1
-6.1
-3.01
-6.01
Figure 3. The graph of h.
-3.001
-6.001
-2.999
-5.999
-2.99
-5.99
-2.9
-5.9
-2.8
-5.8
-2.5
-5.5
Observe that as x approaches  3 , either from the left or from the right, h(x) approaches
 6 or lim g ( x)  6 , eventhough h(3)  6 . Observe too that h is equal to f and g
x  3
except at x  3 .
3
x 9

if x  3
4. Consider the piecewise-defined function t ( x)   x  3
whose graph appears

  6 if x  3
2
in Figure 4.
y-axis
-3
xaxis
-6
x
t (x )
-3.5
-6.5
-3.2
-6.2
-3.1
-6.1
Figure 4. The graph of t.
-3.01
-6.01
-3.001
-6.001
-2.999
-5.999
-2.99
-5.99
-2.9
-5.9
-2.8
-5.8
-2.5
-5.5
Function t behaves exactly the same as function f in the first example and note that we also
lim t ( x )  6 . Function t is what we get when we redefine function g in the second
example by filling the gap at x  3 .
have
x  3
 1 if x  0
5. Consider the function s ( x)  
whose graph is shown in Figure 5.
 1 if x  0
y-axis
1
x-axis
-1
Figure 5. The graph of s.
4
Observe that as x approaches 0 from the left, the function s approaches  1 but as x approaches 0
from the right, s approaches 1. Because of the different function values to which s is approaching
s ( x) does not exist.
as x approaches 0 from both directions, we say that lim
x 0
II. Formal Definition of the Limit of a Function
1. The Definition
Let f be a function that is defined at every number in some open interval containing a number c,
f ( x)  L ,
except possibly at c itself. The limit of f (x) as x approaches c is L, written as lim
x c
if the following statement is true:
Given any   0 , however small, there exists a   0 such that
if 0  x  c   , then f ( x)  L   .
2. Remarks:
f ( x)  L if whenever x assumes a sequence of values that approaches
a. As noted earlier, lim
x c
the number c (both from the left of c and from the right of c), the corresponding sequence of
function values of f approaches L.
b. The number L being the limit of f (x) at x = c can be interpreted geometrically as follows:
for any chosen positive value of  , however small, the function values f (x) on the vertical
axis can be made to lie in the open interval L   , L    by simply letting x on the
horizontal axis be confined in the open interval c   , c    . In other words, if
x  c   , c    , then f (x)  L   , L    . Figure 6 illustrates this.
y-axis
Figure 6
y = f (x)
x-axis
5
c. The number  depends on the chosen  .
d. The limit of a function at a number is unique.
e. It is possible for the limit to exist even if f is not defined at x = c. (See illustration (2) of
Part I.)
f. It is possible for the limit value L to be different from f (c) . (See illustration (3) of Part I.)
3. Illustration:
x2  9
 6 , using
x  3 x  3
Recall the function in illustration (2) of Part I. Prove that lim
  0.005 .
According to the definition, we need to find a
f ( x)  L   . In this particular case, f ( x) 
  0 such that if 0  x  c   , then
x2  9
, L  6 , and c  3 .
x3
First, observe that
f ( x)  L 
x2  9
( x  3)( x  3)
 (6) 
 6  x  3  x  (3)  x  c .
x3
( x  3)
x2  9
 ( 6)    x  (3)   . This suggests that we take
Because of this, it follows that
x3
   , or   0.005. Hence, the existence of  for any positive  chosen.
So, by taking   0.005 , we have indeed 0  x  (3)  0.005 
Note: One may set some other value for  .
III. Basic Limit Theorems
( mx  b)  mc  b .
1. Let m and b constants. Then lim
x c
k k.
2. Let k be a constant. Then lim
x c
3. Let
lim f ( x)  L and lim g ( x )  M . Then
x c
x c
 f ( x)  g ( x)  lim
f ( x)  lim g ( x )  L  M .
a) lim
x c
x c
x c
 f ( x)  g ( x)  lim
f ( x)  lim g ( x)  LM .
b) lim
x c
x c
x c
x2  9
 (6)  0.005 .
x3
6
Note: (a) and (b) can be extended naturally to cases where there are more than two functions.
c) lim
x c
f ( x) L
f ( x) lim
 x c

,M  0 .
g ( x) lim g ( x) M
x c
f ( x)  L and n be a positive integer. Then
4. Let lim
x c


 f ( x)  lim
f ( x)  Ln .
a) lim
x c
x c
n
n
n f ( x )  n lim f ( x )  n L
b) lim
, provided L > 0 when n is even.
x c
x c
f ( x )  f (c ) .
5. If f is a polynomial function, then lim
xc
Exercises: Evaluate the following limits.
( 2 x 3  4 x  5)
a. lim
x2

2
c. lim 4  x
x  1
2 x  3x  1 1  x 2
b. lim
x 1

IV. The Case of the Indeterminate Form
2x 1
x4 x  3
d. lim
(2  x) 5
e. xlim
 3
f. lim
x
4
cos   sin 
csc 2
0
.
0
1. Applying the basic limit theorems of Part III are simply done by direct substitution.
However, direct substitution fails to work when it results into any of these forms:
0 
, ,0  ,00 ,  0 ,1 ,    .
0 
These are called the indeterminate forms.
2. When the evaluation of a limit leads to the indeterminate form 0/0, the limit may or may
not exist. To find this out, we need to remove the indeterminacy by
a) factoring and canceling the factor that zeroes out the numerator and denominator, or
b) rationalizing either the numerator or denominator.
(Dealing with the other indeterminate forms requires techniques beyond the scope of this
lecture.)
7
3.
Example: Consider the function g ( x) 
x 9
g ( x) .
in illustration (2) of Part I. Find xlim
 3
x3
2
Direct substitution of x  3 to g (x) leads to the indeterminate form 0/0. Note that the factor
x  3 zeroes out the numerator and denominator and so must be removed from g by cancellation.
We proceed with the evaluation as follows:
x2  9
( x  3)( x  3)
 lim
 lim ( x  3)  6 .
x  3 x  3
x  3
x  3
( x  3)
lim
4. Example: Find lim
t 0
2 4t
.
t
Direct substitution of t  0 leads to the indeterminate form 0/0. We remove the factor t  0 or t
from the numerator and denominator by rationalizing the numerator. We proceed with the
evaluation as follows:
lim
t 0
2 4t
2 4t 2 4t
4  (4  t )
t
 lim

 lim
 lim
t 0
t 0 t 2  4  t
t
t
2  4  t t 0 t 2  4  t

 lim
t 0



1
1

2 4t 4
5. Exercises: Evaluate the following limits.
x 2  3x  2
x  1
x 1
a. lim
b. lim
x2
x2
x3 8
c. lim
x 3
d. lim
9  x2
x 3
x4
x 4
x 2
e.
f.
x 3  x 2  x  10
x  2
x 2  3x  2
lim
lim
y  3
y3
(1 / y )  (1 / 3)
V. One Sided Limits
1. Right–hand limit or one–sided limit from the right
Let f be a function that is defined at every number in some open interval (a, c). The limit of
f ( x )  L if the following
f (x) , as x approaches a from the right, is L, written as xlim
a 
statement is true: Given any   0 , however small, there exists a   0 such that :
If 0  x  a   , then f ( x)  L   .
2.
Left–hand limit or one–sided limit from the left.
8
Let f be a function that is defined at every number in some open interval (d, a). The limit of
f ( x)  L if the following
f (x) , as x approaches a from the left, is L, written as xlim
a 
statement is true: Given any   0 , however small, there exists a   0 such that :
If 0  a  x   , then f ( x)  L   .
3.
Illustration
s ( x) does not exist. However,
Recall the signum function in Figure 5. As noted earlier, lim
x 0
we can define one-sided limits at x = 0. It can be observed that as x approaches 0 from the left,
the function s approaches  1 , and as x approaches 0 from the right, the function
approaches 1.
s( x)  1 and lim s ( x )  1 .
In symbol, xlim
0 
x 0
4. Remarks:
f ( x)  L will be referred to as two–sided limit.
a) To distinguish from the one–sided limits, lim
xa
b) The basic limit theorems in Pat III remain valid for one–sided limits.
f ( x)  L if and only if lim f ( x)  lim f ( x)  L .
5. Theorem: lim
xa
xa
xa
f ( x )  lim f ( x ) , then lim f ( x) does not exist.
Note: This theorem implies that if xlim
x a
a 
xa
6.
f ( x ) and
Example: Find xlim
2
 x  5 if
5  x if
Since f ( x)  x  5  
lim f ( x ) if f ( x)  x  5 .
x2
x5
and 2 < 5, we take the form f ( x)  5  x . So, we
x5
f ( x )  lim (5  x)  3 and lim f ( x )  lim (5  x)  3 .
have xlim
2
x2
x2
x2
f ( x )  lim f ( x) , we further say that the two-sided limit lim f ( x) exists.
As xlim
x 2
2
x2
7.
r (t ) and lim r (t ) if
Example: Find tlim
4
t 4
t  4 if
r (t )  
4  t if
t4
.
t4
r (t ) , first note that t approaches 4 from the right and so we consider t  4 .
To evaluate tlim
4
r (t )  lim (4  t )  0 .
The form of r(t) we need to take is r (t )  4  t . Hence, tlim
4
t 4
9
r (t ) , we note that t approaches 4 from the left and so we consider t  4 .
To evaluate tlim
4
r (t )  lim (t  4)  8
The form of r(t) we need to take is r (t )  t  4 . Hence, tlim
4
t 4
r (t )  lim r (t ) , we conclude that lim r (t ) does not exist.
As tlim
t 4
4
t 4
8.
Exercises:
Consider the following functions:
i) h( x)  3  2 x  4
 2 if x  1

ii) p ( x)   1 if x  1
 3 if x  1
 3 t  1 if t  1

2
if t  1
iii) g (t )   1  t
 3 t  1 if t  1

Determine whether or not the following limits exist by evaluating appropriate one-sided limits.
h( x )
a) lim
x 2
h( x )
b) xlim
 1
p( x)
c) lim
x1
p ( x)
d) lim
x5
g (t )
e) tlim
 1
g (t )
f) lim
t 1
VI. Infinite Limits
1. Definition
Let f be a function that is defined at every number in some open interval containing a number a,
except possibly at a itself.
f ( x)   if for any
a) As x approaches a, f (x) increases without bound, written as lim
xa
number N > 0, there exists a   0 such that if 0  x  a   , then f ( x)  N .
f ( x)   if for any
b) As x approaches a, f (x) decreases without bound, written as lim
xa
number N > 0, there exists a   0 such that if 0  x  a   , then f ( x)  N .
Figure 7. The graph of y 
1
x2
Figure 8. One-Sided Infinite Limits
10
Figure 7 illustrates an infinite limit of a function y  f (x) at x  0 where the function
increases without bound as x approaches 0 either from the left or from the right. In symbol, we
write lim
x 0
1
  .
x2
One-sided infinite limits may be defined accordingly from (a) and (b). In this case, instead of
x  a , we consider x  a  or x  a  . Figure 8 illustrates this.
2. Remark:
  does not represent large number; it simply tells about the behavior of the function as x
approaches a. So, in both cases, limits do not actually exist.
3. Theorem: If r is any positive integer, then
1
1    if
(a) lim r  
(b) lim r  
x 0 x
x 0 x
  if
r is odd
r is even
f ( x)  0
4. Theorems: If a is any real number and f (x) and g (x) are functions such that lim
xa
g ( x)  c , where c is a nonzero constant, then
and lim
xa
a) If c > 0 and if f (x)  0 through positive values of f (x) , then lim
g ( x)
  .
f ( x)
b) If c > 0 and if f (x)  0 through negative values of f (x) , then lim
g ( x)
  .
f ( x)
c) If c < 0 and if f (x)  0 through positive values of f (x) , then lim
g ( x)
  .
f ( x)
d) If c < 0 and if f (x)  0 through negative values of f (x) , then lim
g ( x)
  .
f ( x)
xa
xa
xa
xa
(Note: The above theorems on infinite limits are also valid for one–sided infinite limits.)
5. Examples
a) Determine if lim
x2
x2
is   or   .
x2  4
b) Discuss the behavior of f ( x) 
1
as x approaches 4.
( x  4) 3
6. Exercises: Indicate either   or   for each of the following limits.
a)
lim
t 0
3t2
t
b)
lim
x 0
4x3  2
5 x 2  3x 3
c)
11
lim  tan 
 
  
2
7. Theorems
f ( x)   and lim g ( x)  c , where c is any constant, then lim f ( x )  g ( x )   .
a) If lim
xa
xa
xa
f ( x)   and lim g ( x)  c , where c is any constant, then lim f ( x )  g ( x )   .
b) If lim
xa
xa
xa
8.
f ( x)   and lim g ( x)  c , where c is a nonzero constant, then
Theorems: If lim
xa
xa
 f ( x)  g ( x)   if c > 0
a) lim
xa
9.
 f ( x)  g ( x)   if c < 0.
b) lim
xa
f ( x)   and lim g ( x)  c , where c is a nonzero constant, then
Theorem: If lim
xa
xa
 f ( x)  g ( x)   if c > 0
a) lim
xa
 f ( x)  g ( x)   if c < 0.
b) lim
xa
8. Remark: Theorems 7 - 9 are also valid for one–sided infinite limits.
9. Exercise: Indicate either   or   for each of the following limits.
 1 2x 1 
 3

x 0 x 4
x 1

a) lim
b)

y 
y 1
lim 

2
y 1 
2 y 1 
 2 y  y 1

10. Vertical Asymptotes
The line x = a is called a vertical asymptote of the graph of the function f (x) if at least one of
the following statements is true:
f ( x)  
a) xlim
a
f ( x)  
b) xlim
a
f ( x)  
c) xlim
a 
f ( x)  
d) xlim
a 
1
whose graph appears in Figure 7. The line x  0
x2
1
serves as a vertical asymptote of the graph of f since lim 2   .
x 0 x
To illustrate, consider the function f ( x) 
12
11. Exercises: Find the equation(s) of vertical asymptote(s) of the graphs of
a) f ( x) 
4
( x  5) 2
b) g (t ) 
t 2 1
t2  2
VII. Limits at Infinity
1. Definition
f (x) be a function that is defined at every number in some interval (a,) . The
f ( x )  L if for any   0 ,
limit of f (x) as x increases without bound is L, written as lim
x 
a)
Let
however small, there exists a number N > 0 such that if x > N, then
b)
f ( x)  L   .
Let f (x) be a function that is defined at every number in some interval (, a) . The limit
f ( x)  L if for any   0 , however
of f (x) as x decreases without bound is L, written as xlim
 
small, there exists a number N > 0 such that if x < N, then
f ( x)  L   .
2. Illustration
As an illustration, recall the function f ( x) 
1
and its graph shown in Figure 7. Observe the
x2
behavior of the function as x tends to positive infinity (or increases without bound) or negative
infinity (decreases without bound). In either case, f approaches a finite value which is 0. In
symbol, we write lim
x 
1
0.
x2
2. Theorem: For any positive integer r, lim
x 
1
1
 lim r  0 .
r
x


x
x
3. Remarks:
a) Basic limit theorems in Part II remain valid for limits at infinity.
b) When evaluating the limit of a rational function p / q at infinity, multiply the numerator
and denominator by 1 / x n , where n is the greater degree between the numerator and
denominator. Moreover, if
(i) degree (p) = degree (q), then the limit is a non-zero constant.
(ii) degree (p) < degree (q), then the limit is zero.
(iii) degree (p) > degree (q), then the limit does not exist.
4x3  2x  5
.
x  5 x 3  2 x 2
4. Example: Evaluate lim
13
3
Since the highest power of x appearing is x , multiply both numerator and denominator by
1 / x 3 . We proceed with the evaluation as follows:
4 x

1
2 5
 2x  5  3 
4 2  3
4x  2x  5
 x   lim
x
x
lim
 lim
x  5 x 3  2 x 2
x 
x


2
1
 
5
5x3  2x 2  3 
x
x 
3
3



1
1
 5 lim 3
2
x


x
x  4  2(0)  5(0)  4
1
5  2(0)
5
lim 5  2 lim
x 
x  x
lim 4  2 lim
x 
x 
5. Exercises: Find the limit if it exists.
7x 2  2x  1
x  3x 2  8 x  5
a) lim
2 x 2  3x  1
x 
5x 3
b) lim
c) lim
x 
4x  1
9x  2
2x7
x  1  5 x 3
d) lim
6. Horizontal Asymptotes
How can the concept of limit at infinity be applied to determine the horizontal asymptote (if any)
of the graph of a function?
Recall f ( x) 
1
1
1
. Since lim 2  lim 2  0 , the line y  0 is the horizontal asymptote of
2
x  x
x   x
x
the graph of f.
7. Exercise: Find the horizontal asymptote of the graph of h( x) 
2x 1
. Sketch the graph of h.
x 1
VIII. CONTINUITY
1. Continuity of a Function at a Number.
A function is said to be continuous at a number x = a if and only if the following conditions are
satisfied:
(a) f (a ) exists,
f ( x) exists,
(b) lim
x a
f ( x)  f ( a)
and (c) lim
xa
If at least one of these conditions are violated, the function is discontinuous at x = a.
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2. Types of Discontinuity at a Number
a) Removable Discontinuity
f ( x) exists but f (a ) does not
A function f has removable discontinuity at x = a if lim
x a
f ( x) exist but lim f ( x)  f (a ) .
exist, or f (a ) and lim
x a
xa
Remark: A function with a removable discontinuity at can be redefined in such a way that it
can be made continuous at the point where its removable discontinuity occurs.
For example, the function g ( x) 
x2  9
in illustration (2) of Part I has a removable
x3
discontinuity at x =  3 . Function t in illustration (4) of Part I is a redefinition of g which
removes the discontinuity of g at x =  3 .
b) Essential Discontinuity
f ( x ) and
A function f has essential discontinuity at x = a if the one-sided limits xlim
a 
lim f ( x) exist but are not equal (implying lim f ( x) does not exist), or when these onex a
xa 
sided limits are infinite.
For example, the signum function in Part I has an essential discontinuity at x = 0 while the
function y 
1
whose graph is shown in Figure 7 has an essential discontinuity at x = 0
x2
too.
3. Some Theorems on Continuity
a) A polynomial function is continuous everywhere.
b) A rational function f (x) is continuous at all values of x except those that make its
denominator 0.
4. Exercises:
a) Determine whether or not
f ( x) 
x2  x  6
is continuous at x = 1. At which point is f
x3
discontinuous? Is this discontinuity removable or essential?
b) Identify the points of discontinuity of the following functions and determine if the
discontinuity is removable or essential.
15
i) f ( x) 
5.
5
x4
t  4 if
4  t if
ii) r (t )  
t4
t4
Continuity on an Interval
A function f is said to be continuous from the right at a number x = a if the following hold:
(a) f (a ) exists,
f ( x ) exists,
(b) xlim
a 
f ( x)  f (a )
and (c) xlim
a 
A function f is said to be continuous from the left at a number x = a if the following hold:
(a) f (a ) exists,
f ( x) exists,
(b) xlim
a 
f ( x)  f ( a)
and (c) xlim
a 
A function f is said to be continuous on an open interval if it is continuous at every point in the
open interval.
A function f whose domain includes the closed interval [a, b] is said to be continuous on [a, b] if
it is continuous on the open interval (a, b), continuous from the right at a, and continuous from
the left at b.
6. Exercises:
a) Determine the interval of continuity of the following functions
(i)
f ( x) 
1
x2
(ii) g ( x) 
x2  9
x3