Chapter 8— Indeterminate Forms and Improper Integrals
Math 1220-004 Class Notes
Section 8.1: Indeterminate Forms of Type
0
0
Fact: The limit of quotient is equal to the quotient of the limits. (book page 68)
Mathematically speaking, this says the following:
lim f (x)
f (x)
= x→a
x→a g(x)
lim g(x)
lim
x→a
Provided that the denominator is non-zero.
However, this is unable to determine the limits of the type
Example 1. A classic examples to consider is :
lim
x→0
1 − cos x
x
1
0
.
0
However, we have the following powerful tool, which is the L’Hôpital’s Rule.
L’Hôpital’s Rule
Suppose that lim f (x) = lim g(x) = 0. Then:
x→u
x→u
f 0 (x)
f (x)
= lim 0
x→u g (x)
x→u g(x)
lim
Verbally speaking, this says that the quotient limit of two functions is equal to the
quotient limit of their derivatives, provided that both of them are originally approaching
0 at the same time.
Example 1 (Revisited)
(a)
(b)
sin x
x
1 − cos x
x
2
Example 2. Calculate the following limits
(1)
lim
x→0
tan 2014x
ln (1 + 801x)
(2)
lim
x→0
sin x − x
x3
3
(3)
1 − cos x
x→0 x2 + 3x
lim
4
(4) (a)
Rx√
0
lim
x→0
(b)
t sin tdt
x2
R x2 √
lim
0
x→0
5
t cos tdt
x2
(5)
ex − ln (1 + x) − 1
x→0
x2
lim
(6)
lim
x→0
cosh x − 1
x2
6
Section 8.2: Other Indeterminate Forms to Consider
0
∞
. How about the type
?
0
∞
x2
x2
x2
, lim
and
lim
Consider lim
x→∞ x x→∞ 2x2
x→∞ 2x3
We have considered the type
Luckily enough, L’HÔpital’s Rule works for the type
∞
as well.
∞
L’Hôpital’s Rule
Suppose that | lim f (x)| = | lim g(x)| = ∞. Then:
x→u
x→u
f (x)
f 0 (x)
= lim 0
x→u g(x)
x→u g (x)
lim
where u could be any of the following: a, a+ ,a− ,−∞, ∞.
Verbally speaking, this says that the quotient limit of two functions is equal to the
quotient limit of their derivatives, provided that both of them are originally approaching
infinity (either positive or negative) at the same time.
7
Examples of type
0
.
0
(1) (a) Show that for any positive number n,
ex
=∞
x→∞ xn
lim
(b) Show that for any positive number n,
lim
x→∞
8
ln x
=0
xn
(2)
lim
x→0+
ln x
cot x
(3)
lim (tan x · ln (sin x))
x→ π2
9
(4)
lim (
x→1+
1
x
−
)
x − 1 ln x
A list of Non-Indeterminate forms
Forms
Results
0
∞
∞
0
∞ (positive)+ ∞ (positive)
∞·∞
0∞
∞∞
10
10
A list of indeterminate forms (whose results can not be determined solely by their
expressions):
(1)
0
0
(2)
±∞
±∞
(3) 0 ∗ ∞
(4) ∞ − ∞
(5) 00
(6) ∞0
(7) 1∞
We now deal with three last types 00 , ∞0 , 1∞
Strategy: Write everything in their exponential forms.
Examples
(1)
lim (x + 1)cot x
x→0+
11
(2)
lim (tan x)cos x
x→ π2 −
12
Section 8.3: Improper Integrals: Infinite Limits of Integration
Rb
The integral a f (x)dx measures the area under the curve y = f (x) between a and b,
where we have assumed that both a and b are finite numbers.
Suppose, however, you want to measure the area of a curve between a certain number
a and positive infinity ∞, then the first thing you might try is to calculate the integral
of that function between a and some arbitrary number b (which you know how to solve
already) and then let b go to infinity, if appropriate.
For example, to compute
R∞
e−x dx, we do the following:
0
The above example actually, indeed, illustrates the way we both define and compute
an improper integral.
Definition
Z
Z
b
b
f (x)dx = lim
−∞
Z
Z
∞
f (x)dx = lim
a
f (x)dx
a→−∞ a
b→∞ a
b
f (x)dx
If the limits on the right exist and have finite values, then we say that the corresponding improper integrals converge and have those values. Otherwise, the integrals
are said to diverge.
In sum, two steps to calculate an improper integral:
(1)
(2)
13
Examples
(1)
Z
∞
1
(2)
Z
2
dx
√
πx
∞
x2
14
1
−x
(3)
Z
2
e4x dx
−∞
(4)
Z
1
−∞
dx
(3x − 4)5
15
Definition
If both
Z
0
f (x)dx
−∞
and
Z
∞
f (x)dx
0
exist, then we define
Z
Z
∞
Z
0
f (x)dx =
−∞
Z
f (x)dx +
−∞
f (x)dx
0
∞
and we say that
f (x)dx converges. Otherwise, diverges.
−∞
Example
∞
Z
∞
−∞
1
dx
1 + x2
16
Section 8.4: Improper Integrals: Infinite Integrands
Consider the integral:
Z
1
−2
¯1
1
1 ¯¯
3
1
dx = [− ]¯ = −1 − = −
x2
x −2
2
2
1
is always positive so that we should expect
x2
3
the result to be greater than zero, however, − < 0.
2
Clearly this is wrong. The integrand
Reason: for a function to be integrable in the proper sense it must be bounded. How1
ever, the function 2 is unbounded when x → 0.
x
Definition
Let f be continuous on the half-open interval [a, b) and suppose that lim |f (x)|dx = ∞.
Then
x→b−
Z
Z
b
t
f (x)dx = lim
a
t→b−
f (x)dx
a
provided that this limit exists and is finite, in which case we say that the integral
converges. Otherwise, we say that the integral diverges.
Similarly, let f be continuous on the half-open interval (a, b] and suppose that
lim |f (x)|dx = ∞. Then
x→a+
Z
Z
b
f (x)dx = lim
a
t→a+
b
f (x)dx
t
provided that this limit exists and is finite, in which case we say that the integral
converges. Otherwise, we say that the integral diverges.
17
Examples
(1)
Z
4
√
0
(2) Show that
Z
dx
16 − x2
1
0
converges if p < 1, but diverges if p ≥ 1.
18
1
dx
xp
(3)
Z π
4
sec2 x
dx
(tan x − 1)2
0
Definition
Let f be continuous on the interval [a, b] except at a number c, where a < c < b, and
suppose that lim |f (x)|dx = ∞. Then we define
x→c
Z
Z
b
f (x)dx =
a
Z
c
f (x)dx +
a
b
f (x)dx
c
provided that this limit exists and is finite, in which case we say that the integral
converges. Otherwise, we say that the integral diverges.
19
Examples
(1)
Z
0
5
x
dx
9 − x2
20