Evaluating Limits
Limits can be evaluated
I
numerically
I
using Limit Laws to reduce the limit to a known limit
I
algebraically
I
using the Squeeze Theorem
We will cover the first two of these approaches in this lecture and other two
in the next lecture.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
2/23
Evaluating Limits Numerically
When you have a formula for the function f , the value of the limit
lim f (x )
x →a
can be computed numerically by computing the value of f at a sequence of
x values that approaches the limit point a. For a regular, two-sided, limit
this must be done for two sequences of x values approaching a, one from
above (the right) and one from below (the left). This is best done using a
table of values.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
3/23
Example 6 – A Trigonometric Limit
Consider the limit
1 − cos x
lim
x →0
x2
1 − cos x
x2
x
Note that substitution of x = 0 into
1 − cos x
gives
the function f (x ) =
x2
0
. Use a
0
table of values to estimate the value
of the limit.
−0.1000
−0.0100
−0.0010
−0.0001
0
the indeterminate form
?
0.0001
0.0010
0.0100
0.1000
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
4/23
Continuing Example 6
Note that the values in the upper half of the table duplicate those in the
lower half. Why? Recall that a function f is even if
f (−x ) = f (x )
In this case the graph of the function is symmetric across the y -axis. Both
g (x ) = cos x and h (x ) = x 2 are even. Their graphs are
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
5/23
Continuing Example 6
Thus
f (−x ) =
Further, recall that a function f is odd if f (−x ) = −f (x ). The graph of an
odd function is symmetric through the origin. Also, for an odd function you
must have f (0) = 0, or
lim f (x ) = 0
x →0
You convince yourself that sin x, x 3 , x 3 cos x, and x 2 sin x are odd
functions, but that x 3 sin x is even.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
6/23
Continuing Example 6
What if we get carried away with our
calculation and add a few more steps
to the table?
x
1 − cos x
x2
0.0000100
0.5
0.0000010
0.5
0.0000001
1.0
OOPS!! Are we wrong about the value of the limit? No!! This is an artifact of your calculator.
You may run into problems for a different x value and you may see a different value for the
function. What is happening here is that your calculator rounds any calculation to a certain
number of digits. In a calculation like the limit above in which one value almost exactly cancels
another, as in the numerator of the function above, this rounding eventually gives incorrect
values. Watch out for this when you estimate limits numerically. You see what happens with
the limit
1 − 12 x 2 − cos x
lim
x →0
x4
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
7/23
Example 7 – Limit of an Exponential Function
Consider the limit
x
2x − 2
lim
x →1 x − 1
0.500
Again note that substitution of x = 1
0
gives . Estimate the value of the
0
limit using two sequences of values:
x = 1.5, 1.1, 1.01, 1.001 and
x = 0.5, 0.9, 0.99, 0.999. The
values do seem to be approaching a
limiting value, but there is not a nice
round value.
Clint Lee
2x − 2
x −1
0.900
0.990
0.999
1.000
?
1.001
1.010
1.100
1.500
Math 112 Lecture 2: Evaluation of Limits I
8/23
Continuing Example 7
We can get a more accurate estimate by averaging the two closest values.
This gives an improved estimate of
1.3867749252 + 1.3858140191
= 1.38629447215
2
This is still not pretty.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
9/23
Some Limit Laws
Once you have evaluated the limit of one or more functions at a specific
limit point, you can use Limit Laws to evaluate other limits involving the
same functions at that limit point. Suppose that
lim f (x )
x →a
and
lim g (x )
x →a
exist. Then some examples of Limit Laws are (a complete list of Limit Laws
is given in Section 2.3, pages 99 - 101 of your text):
lim [f (x ) + g (x )] = lim f (x ) + lim g (x )
1.
x →a
3.
x →a
lim [cf (x )] = c lim f (x )
x →a
5.
lim
x →a
7.
x →a
lim f (x )
f (x )
x →a
=
g (x )
lim g (x )
x →a
lim c = c
x →a
where c is a constant
if lim g (x ) 6= 0
x →a
where c is a constant
x →a
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
10/23
Comments on Limit Laws
Note that in many cases the Limit Laws give rules that allow you to
interchange certain mathematical operations with limits.
I
Rule 1 states that you can interchange taking a limit with addition.
I
Rule 5 states that you can interchange taking a limit with division.
I
Exceptions are Rules 3 and 7, which deal with the behavior of limits
when there are constants.
Note that Rule 3 can be derived from Rules 4 and 7. Since
lim [cf (x )] =
x →a
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
11/23
Another Limit Law
A Limit Law dealing with the root of a function is
11.
lim
x →a
q
n
f (x ) =
q
n
lim f (x )
x →a
where n is a positive integer
Note that if n is even we must assume that lim f (x ) > 0. Otherwise, it is
x →a
possible that the limit does not exist.
This is because, for example, the limit lim
lim
x →0+
√
x →0
x =0
but
√
x
lim
x →0−
DNE, since
√
x DNE
You can’t take the square root of number less than zero. This is not an
issue if f (x ) ≥ 0 for all x on both sides of a, such as f (x ) = |x |.
This is not a problem for n odd, since you can take an odd root of any real
number, positive or negative or zero.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
12/23
Example 8 – Applying the Limit Laws
Consider the function
f (x ) =
s
x 2 − 16
6−x
(a) Determine the domain of the function f .
(b) Evaluate the limit lim f (x ), or explain why the limit does not exist.
x →5
(c) Evaluate the limit lim f (x ), or explain why the limit does not exits.
x →4
(d) Evaluate the limit lim f (x ), or explain why the limit does not exits.
x →7
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
13/23
Solution: Example 8(a)
In order to be able to take the square root, we must have
x 2 − 16
≥0
6−x
There are only three x values where this expression can change sign:
x = 4, x = −4, or x = 6, which partition the number line into four intervals:
(−∞, −4), (−4, 4), (4, 6), and (6, ∞). Choose a test value in each
interval and determine the sign there.
Interval
Test value
Sign
(−∞, −4)
x = −5
+
Clint Lee
(−4, 4)
(4, 6)
(6, ∞)
Math 112 Lecture 2: Evaluation of Limits I
14/23
Solution: Example 8(a) – continued
The table can be summarized on the number line below
−4
4
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
6
15/23
Solution: Example 8(b)
First note that using Rules 2, 7, 8, and 9 gives
lim
x →5
x 2 − 16 =
lim (6 − x ) =
x →5
Then, since the second limit is not zero, we can use Rule 5 to give
x 2 − 16
=
x →5 6 − x
lim
So, finally, since this limit is positive, we can use Rule 11 to give
lim
x →5
s
x 2 − 16
=
6−x
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
16/23
The Direct Substitution Property
The solution to Example 8(b) just given demonstrates that in many cases a
limit can be evaluated by direct substitution. That is
lim f (x ) = f (a)
x →a
The Direct Substitution Property on page 102 of the text states that this
works for all polynomial and rational functions, unless the denominator of
the rational function is zero at the limit point. This is illustrated in the first
two steps in the solution just given, which demonstrate the basic ideas that
are used in the proof of the general result. In fact, the complete solution
shows that the Direct Substitution Property works in more general cases as
well, in particular for roots of polynomial or rational functions.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
17/23
Polynomial Functions
If n is a nonnegative integer, then a polynomial function p of degree n is of
the form
p ( x ) = a 0 + a1 x + a 2 x 2 + · · · + a n x n
where the constants a0 , a1 , a2 , . . . , an are the coefficients of the
polynomial. The domain of a polynomial function is all real numbers.
Note:
I
A constant function, p (x ) = a0 , is a zero degree polynomial.
I
A linear function, p (x ) = a0 + a1 x, is a first degree polynomial.
I
A quadratic function, p (x ) = a0 + a1 x + a2 x 2 , is a second degree
polynomial.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
18/23
Rational and Algebraic Functions
If p and q are polynomial functions, then a rational function is of the form
r (x ) =
p (x )
q (x )
The domain of a rational function is all x where the denominator is not zero,
i.e., q (x ) 6= 0. Note that a polynomial function is a rational function with
q (x ) = 1.
An algebraic function is a combination of rational (or polynomial) functions
that can involve roots. The domain of an algebraic function is all x where
any denominators are nonzero and any quantity under an even root is
nonnegative. A rational (or polynomial) function is an algebraic function with
no roots.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
19/23
Direct Substitution in Algebraic Functions
From our discussion of the Direct Substitution Property, we see that Direct
Substitution can be used in limits involving algebraic functions (includes
rational and polynomial functions) unless
I
I
I
Substitution gives a zero in a denominator (not in domain).
0
Check for the indeterminate form .
0
Substitution gives a negative number under an even root (not in
domain).
Substitution gives zero under an even root (possibly in domain).
The limit may or may not exist. It is necessary to investigate the
one-sided limits.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
20/23
Solution: Example 8(c)
Using the same steps as in the solution to part (b) we see that
x 2 − 16
lim
=0
x →4 6 − x
so we cannot use Rule 11. However, this does not guarantee that the limit
does not exist. We must check the two one-sided limits at x = 4 to
determine if the limit exists or not. Note that for −4 < x < 4, f (x ) < 0.
Thus,
s
s
lim
x →4−
x 2 − 16
6−x
⇒ lim
x →4
Clint Lee
x 2 − 16
6−x
Math 112 Lecture 2: Evaluation of Limits I
21/23
Solution: Example 8(c) – continued
Note that the limit
lim
x →4
does exist since
s
2
x − 16
|x 2 − 16|
6−x
≥ 0 for x 6= 6, which means that the both
6−x
one-sided limits of the square root exist and equal zero.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
22/23
Solution: Example 8(d)
Now
x 2 − 16
= −33 < 0
x →7 6 − x
lim
Since we cannot take the square root of a negative number, the limit of f at
x = 7 does not exist.
Clint Lee
Math 112 Lecture 2: Evaluation of Limits I
23/23