There is an interesting rule which is often useful for evaluating difficult limits called
L’Hôpital’s^{1}
rule. The best versions of this rule are based on the Cauchy Mean value theorem, Theorem
7.8.2 on Page 386.
Theorem 8.6.1Let
[a,b]
⊆
[− ∞, ∞ ]
and suppose f,g are functionswhichsatisfy,
lim f (x) = lim g(x) = 0, (8.18)
x→b− x→b−
(8.18)
and f^{′}and g^{′}exist on
(a,b)
with g^{′}
(x)
≠0 on
(a,b)
. Suppose also that
f′(x)
xli→mb− g′(x ) = L. (8.19)
(8.19)
Then
f (x)
xl→imb− g(x)-= L. (8.20)
(8.20)
Proof: By the definition of limit and 8.19 there exists c < b such that if t > c,
then
|| ′ ||
||f′(t)-− L || < ε.
g (t) 2
Now pick x,y such that c < x < y < b. By the Cauchy mean value theorem, there exists
t ∈
The following corollary is proved in the same way.
Corollary 8.6.2Let
[a,b]
⊆
[− ∞, ∞]
and suppose f,g are functions whichsatisfy,
xli→ma+ f (x) = xl→ima+ g(x) = 0, (8.21)
(8.21)
and f^{′}and g^{′}exist on
(a,b)
with g^{′}
(x)
≠0 on
(a,b)
. Suppose also that
lim f′(x)= L. (8.22)
x→a+ g′(x)
(8.22)
Then
lim f-(x)-= L. (8.23)
x→a+ g(x)
(8.23)
Here is a simple example which illustrates the use of this rule.
Example 8.6.3Find lim_{x→0}
5x+sin3x
tan7x
.
The conditions of L’Hôpital’s rule are satisfied because the numerator and denominator
both converge to 0 and the derivative of the denominator is nonzero for x close to 0.
Therefore, if the limit of the quotient of the derivatives exists, it will equal the limit of the
original function. Thus,
Warning 8.6.5Be sure to check the assumptions of L’Hôpital’srule before using it.
Example 8.6.6Find lim_{x→0+}
cos2x
x
.
The numerator becomes close to 1 and the denominator gets close to 0. Therefore, the
assumptions of L’Hôpital’s rule do not hold and so it does not apply. In fact there is no
limit unless you define the limit to equal +∞. Now lets try to use the conclusion of
L’Hôpital’s rule even though the conditions for using this rule are not verified.
Take the derivative of the numerator and the denominator which yields
−2sin2x
1
, an
expression whose limit as x → 0+ equals 0. This is a good illustration of the above
warning.
Some people get the unfortunate idea that one can find limits by doing experiments with a
calculator. If the limit is taken as x gets close to 0, these people think one can find the
limit by evaluating the function at values of x which are closer and closer to 0.
Theoretically, this should work although you have no way of knowing how small you need
to take x to get a good estimate of the limit. In practice, the procedure may fail
miserably.
Example 8.6.7Find lim_{x→0}
10
ln|1x+1x0-|
.
This limit equals lim_{y→0}
ln|1y+y|
= lim_{y→0}
(-1-)
-1+1y-
= 1 where L’Hôpital’s rule has been used.
This is an amusing example. You should plug .001 in to the function
ln|1+x10|
x10
and see what
your calculator or computer gives you. If it is like mine, it will give 0 and will keep on
returning the answer of 0 for smaller numbers than .001. This illustrates the folly
of trying to compute limits through calculator or computer experiments. Indeed,
you could say that a calculator is as useful for understanding limits as a bicycle
is for swimming. Those who pretend otherwise are either guilty of ignorance or
dishonesty.
There is another form of L’Hôpital’s rule in which lim_{x→b−}f
(x )
= ±∞ and
lim_{x→b−}g
(x)
= ±∞.
Theorem 8.6.8Let
[a,b]
⊆
[− ∞, ∞ ]
and suppose f,g are functionswhichsatisfy,
xli→mb− f (x) = ±∞ and xli→mb− g(x) = ±∞, (8.24)
(8.24)
and f^{′}and g^{′}exist on
(a,b)
with g^{′}
(x)
≠0 on
(a,b)
. Suppose also
f′(x)
xli→mb− g′(x-) = L. (8.25)
(8.25)
Then
f (x)
xl→imb− g(x)-= L. (8.26)
(8.26)
Proof: By the definition of limit and 8.25 there exists c < b such that if t > c,
then
| |
||f′(t)-− L ||< ε.
|g′(t) | 2
Now pick x,y such that c < x < y < b. By the Cauchy mean value theorem, there exists
t ∈
As before, there is no essential difference between the proof in the case where x → b− and
the proof when x → a+. This observation is stated as the next corollary.
Corollary 8.6.9Let
[a,b]
⊆
[− ∞,∞ ]
and suppose f,g are functions whichsatisfy,
lx→iam+ f (x) = ±∞ and xli→ma+ g(x) = ± ∞, (8.27)
(8.27)
and f^{′}and g^{′}exist on
(a,b)
with g^{′}
(x)
≠0 on
(a,b)
. Suppose also that
f′(x)
xli→ma+ g′(x) = L. (8.28)
(8.28)
Then
lim f-(x)-= L. (8.29)
x→a+ g(x)
(8.29)
Theorems 8.6.18.6.8 and Corollaries 8.6.2 and 8.6.9 will be referred to as L’Hôpital’s rule
from now on. Theorem 8.6.1 and Corollary 8.6.2 involve the notion of indeterminate forms of
the form
0
0
. Please do not think any meaning is being assigned to the nonsense expression
0
0
.
It is just a symbol to help remember the sort of thing described by Theorem 8.6.1 and
Corollary 8.6.2. Theorem 8.6.8 and Corollary 8.6.9 deal with indeterminate forms which are of
the form
±-∞
∞
. Again, this is just a symbol which is helpful in remembering the
sort of thing being considered. There are other indeterminate forms which can be
reduced to these forms just discussed. Don’t ever try to assign meaning to such
symbols.
Example 8.6.10Find lim_{y→∞}
( )
1 + x
y
^{y}.
It is good to first see why this is called an indeterminate form. One might think that as
y →∞, it follows x∕y → 0 and so 1 +
x
y
→ 1. Now 1 raised to anything is 1 and so it would
seem this limit should equal 1. On the other hand, if x > 0, 1 +
x
y
> 1 and a number
raised to higher and higher powers should approach ∞. It really isn’t clear what this
limit should be. It is an indeterminate form which can be described as 1^{∞}. By
definition,