The caught mean value theorem and L'Hopital's rule are included in this module, accompanied by practice exercises.
Many limits of certain combinations of functions are difficult to evaluate
because they lead to what's known as“indeterminate forms.” These are expressions of the form
and the like.
They are precisely combinations of functions that are not covered by our limit theorems. See
[link] .
The very definition of the derivative itself is such a case:
and we
are interested in the limit of the quotient of these two functions, which would lead us to the indeterminate form
The definition of the number
is another example:
and we are interested in the limit of
which leads to the indeterminate form
L'Hopital's Rule,
[link] below, is our strongest tool for handling such indeterminate forms.
To begin with, here is a useful generalization of the Mean Value Theorem.
Cauchy mean value theorem
Let
and
be continuous real-valued functions on a closed interval
suppose
and assume that both
and
are differentiable on the open interval
Then there exists a point
such that
Prove the preceding theorem.
HINT: Define an auxiliary function
as was
done in the proof of the original Mean Value Theorem.
The following theorem and exercise comprise what is
called L'Hopital's Rule.
Suppose
and
are differentiable real-valued functions
on the bounded open interval
and assume that
where
is a real number.
(Implicit in this hypothesis is that
for
in some interval
)
Suppose further that either
or
then
Suppose first that
Observe first that, because
for all
in some interval
is either always positive or always negative on that interval.
(This follows from part (d) of
[link] .)
Thereforethe function
must be strictly monotonic on the interval
Hence, since
we must have that
on the interval
Now, given an
choose
a positive
such that
if
then
Then, for every natural number
for which
and every
we have by the Cauchy Mean Value Theorem that
there exists a point
between
and
such that
Therefore, taking the limit as
approaches
we obtain
for all
for which
This proves the theorem in this first case.
Next, suppose that
This part of the theorem is a bit more complicated to prove.
First, choose a positive
so that
and
are both
positive on the interval
This is possible because both functions are tending to infinity as
approaches
Now, given an
choose a positive number
such that
for all
We express this absolute value inequality as the following pair of ordinary inequalities:
Set
Using the Cauchy Mean Value Theorem, and the preceding inequalities, we
have that for all
implying that
Dividing through by
and simplifying we obtain
Finally, using the hypothesis that
and the fact that
and
are all constants,
choose a
with
such that if
then
and
Then, for all
we would have
implying that
and the theorem is proved.
- Show that the conclusions of the preceding theorem also hold if we assume that
HINT: Replace
by a large real number
and show that
if
- Show that the preceding theorem, as well as part (a) of this exercise,
also holds if we replacethe (finite) endpoint
by
HINT: Replace the
's by negative numbers
- Show that the preceding theorem, as well as parts a and b of this exercise,
hold if the limit as
approaches
from the right
is replaced by the limit as
approaches
from the left.
HINT: Replace
by
and
by
- Give an example to show that the converse
of L'Hopital's Rule need not hold; i.e., find functions
and
for which
- Deduce from the proof given above that if
and
then
independent of the behavior of
- Evaluate
and
HINT: Take logarithms.