Download Limits of Functions

Limits of Functions
x 1
Let's examine the function f  x  =
x
What happens to f(x) as x increases?
x 1
we write lim
1
x 
x
x 1
and we say "the limit of
as x approaches infinity is 1"
x
Because f(x) is defined for negative as well as positive numbers, we can
also talk about “the limit of f(x) as x approaches negative infinity.”
x 1
lim
1
x 
x
x 1
These symbols mean that
can be made as close to 1 as we like just by
x
considering negative values of x with large enough absolute value. If x  1,000,000
x  1 999,999

1
x
1,000,000
A function f for which f  x  becomes arbitrarily large as x approaches  or  
has no finite limit.
It instead has an infinite limit, either  or  .
Although we say that an infinite limit exists, we recognize that  and  
are not numbers and so we must define finite and infinite limits differently
The following examples should help clarify the concept of an infinite
limit.
Example 1. Evaluate: a. lim x
1
3
x
b. lim x
1
3
1
3
x
a. Since the value of x becomes arbitrarily large as x becomes arbitrarily large,
1
3
lim x  
x 
1
3
b. When x is negative, the value of x is negative.
As x approaches negative infinity x
1
3
becomes arbitrarily large
1
3
lim x  
x
Example 2. Explain why lim x sin x  
x
The Limit as x Approaches a Real Number c
To determine the behavior of a function f(x) as x approaches a real
number c, we consider the following two limits:
f  x  , read "the limit of f  x  as x approaches c from the right."
1 xlim
c

f  x  , read "the limit of f  x  as x approaches c from the left."
 2 xlim
c

Example 3. Using the graph of f  x  shown at the
right, find lim f  x  and lim f  x  .
x 2
x 2
The fact that f  2  3 has nothing to do with the solution.
When evaluating lim f  x  , we are concerned with the value of f  x  for x
x 2
near, but greater than, 2.
Since the value of f  x  gets closer and closer to  2 as x approaches 2 from the right
lim f  x   2
x2
Likewise, the value of f  x  gets closer and closer to 5 as x approaches 2 from the left
lim f  x   5
x2
x2  4
Example 4. If f  x  
, describe the behavior of f  x  near x  2.
x2
The fact that f(2) is undefined has nothing to do with the solution.
The problem is to determine whether the value of f(x) gets close to
any number as x gets closer to 2.
By substituting values of x near 2, we get the values of f(x) below
f  2.1  4.1
f 1.9   3.9
f  2.01  4.01
f 1.99   3.99
f  2.001  4.001
f 1.999   3.999
lim f  x   4
x2
lim f  x   4
x2
Notice in example 4 that lim f  x   lim f  x   4
x2
x2
In this case, we can speak of "the limit of f  x  as x approaches 2" and write
lim f  x   4
x2
In example 3, however, lim f  x   2 and lim f  x   5
x2
x2
Since the right-hand and left-hand limits are different, lim f  x  does not exist.
x2
lim f  x  exists if and only if lim f  x  and lim f  x  exist and agree.
xc
xc
xc
Continuous Functions
Often the easiest limits to evaluate are those
involving a continuous function.
A function is continuous if you can draw
its graph without lifting your pencil from
the paper.
For example, f  x   x3  3x2  3x  1 at right
A function f  x  is continuous at a real number c if: lim f  x   f  c 
xc
In this definition note that there are three conditions for continuity at x = c
1. lim f  x  must exist.
xc
2. f  c  must exist.
3. 1 and  2 must be equal.
A function can fail to be continuous at x = c in different ways.
lim f  x  does not exist
xc
f  c  does not exist
lim f  x   f  c 
xc
Quotient Theorem for Limits
If lim f(x) and lim g(x) both exist, and lim g(x)  0, then
f  x  lim f  x 
lim

g  x  lim g  x 
x 1
Example 5. Evaluate lim 2
x3 x  1
x 1
2 1
x  1 lim
4
x3

lim 2



2
x3 x  1
lim  x  1
8
4
8
x3
Techniques for Evaluating
f  x
lim
g  x
1. If possible, use the quotient theorem for limits.
2. If lim f(x) = 0 and lim g(x) = 0, try the following techniques.
f  x
a. Factor g  x  and f  x  and reduce
to lowest terms.
g  x
b. If f  x  or g  x  involves a square root, try multiplying both
f  x  and g  x  by the conjugate of the square root expression.
3. If lim f  x   0 and lim g  x   0, then either statement  a  or b  is true.
a
f  x
lim
does not exist.
g  x
 b  lim
f  x
f  x
  or lim
 
g  x
g  x
4. If x is approaching infinity or negative infinity, divide the numerator
and the denominator by the highest power of x in the denominator.
f  x
f  x
5. If all else fails, you can guess lim
by evaluating
x g  x 
g  x
f  x
f  x
by evaluating
xc g  x 
g  x
for values of x very near x  c. This was done in example 4.
for very large values of x. Also, you can guess lim
x2  4
Example 6. Evaluate: a. lim
x 2 x  2
1 1 x
b. lim
x0
x
a. Use technique 2a.
x2  4
x  2  x  2 

lim
= lim
=
lim
x

2


x 2 x  2
x2
x2
x2
b. Use technique 2b.
1 1 x
lim
x 0
x
=4
1
2


1 x
1  1  x 1  1  x  lim
 lim

x0
x
1  1  x x0 x 1  1  x
x
 lim
x 0
x 1 1 x


1
 lim
x0 1  1  x


2

1

2
Example 7. Evaluate: a. lim
x1
1
 x  1
2
1
b. lim
x1 x  1
x3  4 x
Example 8. Evaluate lim 4
x 2 x  5
Use technique 4
divide the numerator and denominator by x 4
1 4
 3
00
x3  4 x
x
x

 lim

0
lim 4
x
20
5
x 2 x  5
2 4
x
Formal Definitions of Limits
1. lim f  x   L means that for any small positive
xc
number   epsilon  , there is a positive
number   delta  such that
f  x  L  
whenever x is in the domain of f
and 0 < x  c   .
2. lim f  x   L means that for any small positive
x
number  there is a value of x, call it x1
such that
f  x  L  
whenever x is in the domain of f and x  x1.
3. lim f  x    means that for any large positive
x
number M , there is a value of x, call it x1, such that
f  x  M
whenever x is in the domain of f and x  x1.