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Indeterminate Forms and L’Hospital’s Rule
Back in the chapter on Limits we saw methods for dealing
with the following limits.
In the first limit if we plugged in 
we would get 0/0 and in the second limit if we
“plugged” in infinity we would get (recall that as x goes to infinity a polynomial will
behave in the same fashion that it’s largest power behaves). Both of these are called indeterminate forms. In both
of these cases there are competing interests or rules and it’s not clear which
will win out.
In the case of 0/0 we typically think of a fraction that has
a numerator of zero as being zero.
However, we also tend to think of fractions in which the denominator is
going to zero as infinity or might not exist at all. Likewise, we tend to think of a fraction in
which the numerator and denominator are the same as one. So, which will win out? Or will neither win out and they all “cancel
out” and the limit will reach some other value?
In the case of we have a similar set of problems. If the numerator of a fraction is going to
infinity we tend to think of the whole fraction going to infinity. Also if the denominator is going to infinity
we tend to think of the fraction as going to zero. We also have the case of a fraction in which
the numerator and denominator are the same (ignoring the minus sign) and so we
might get -1. Again, it’s not clear
which of these will win out, if any of them will win out.
With the second limit there is the further problem that
infinity isn’t really a number and so we really shouldn’t even treat it like a
number. Much of the time it simply won’t
behave as we would expect it to if it was a number. To look a little more into this check out the
Types of Infinity section in the Extras
chapter at the end of this document.
This is the problem with indeterminate forms. It’s just not clear what is happening in the
limit. There are other types of indeterminate
forms as well. Some other types are,
These all have competing interests or rules that tell us
what should happen and it’s just not clear which, if any, of the interests or
rules will win out. The topic of this section
is how to deal with these kinds of limits.
As already pointed out we do know how to deal with some
kinds of indeterminate forms already.
For the two limits above we work them as follows.
In the first case we simply factored, canceled and took the
limit and in the second case we factored out an from both the numerator and the
denominator and took the limit. Notice
as well that none of the competing interests or rules in these cases won
out! That is often the case.
So we can deal with some of these. However what about the following two limits.
This first is a 0/0 indeterminate form, but we can’t factor
this one. The second is an indeterminate form, but we can’t just factor
an out of the numerator. So, nothing that we’ve got in our bag of
tricks will work with these two limits.
This is where the subject of this section comes into
play.
L’Hospital’s Rule
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Suppose that we have one of the following cases,
where a can be
any real number, infinity or negative infinity. In these cases we have,
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So, L’Hospital’s Rule tells us that if we have an
indeterminate form 0/0 or all we need to do is differentiate the
numerator and differentiate the denominator and then take the limit.
Before proceeding with examples let me address the spelling
of “L’Hospital”. The more correct
spelling is “L’Hôpital”. However, when I
first learned Calculus I my teacher used the spelling that I use in these notes
and the first text book that I taught Calculus out of also used the spelling
that I use here. So, I’m used to
spelling it that way and that is the way that I’ve spelled it here.
Let’s work some examples.
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Example 1 Evaluate
each of the following limits.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
So, we have already established that this is a 0/0
indeterminate form so let’s just apply L’Hospital’s Rule.
[Return to Problems]
(b)
In this case we also have a 0/0 indeterminate form and if
we were really good at factoring we could factor the numerator and
denominator, simplify and take the limit.
However, that’s going to be more work than just using L’Hospital’s
Rule.
[Return to Problems]
(c)
This was the other limit that we started off looking at
and we know that it’s the indeterminate form so let’s apply L’Hospital’s Rule.
Now we have a small problem. This new limit is also a indeterminate form. However, it’s not really
a problem. We know how to deal with
these kinds of limits. Just apply
L’Hospital’s Rule.
Sometimes we will need to apply L’Hospital’s Rule more
than once.
[Return to Problems]
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L’Hospital’s Rule works great on the two indeterminate
forms 0/0 and . However, there are many more indeterminate
forms out there as we saw earlier. Let’s
take a look at some of those and see how we deal with those kinds of
indeterminate forms.
We’ll start with the indeterminate form .
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Example 2 Evaluate
the following limit.
Solution
Note that we really do need to do the right-hand limit
here. We know that the natural
logarithm only defined for positive x
and so this is the only limit that makes any sense.
Now, in the limit, we get the indeterminate form . L’Hospital’s Rule won’t work on products,
it only works on quotients. However,
we can turn this into a fraction if we rewrite things a little.
The function is the same, just rewritten, and the limit is
now in the form and we can now use L’Hospital’s Rule.
Now, this is a mess, but it cleans up nicely.
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In the previous example we used the fact that we can always
write a product of functions as a quotient by doing one of the following.
Using these two facts will allow us to turn any limit in the
form into a limit in the form 0/0 or . One of these two we get after doing the
rewrite will depend upon which fact we used to do the rewrite. One of the rewrites will give 0/0 and the
other will give . It all depends on which function stays in the
numerator and which gets moved down to the denominator.
Let’s take a look at another example.
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Example 3 Evaluate
the following limit.
Solution
So, it’s in the form . This means that we’ll need to write it as a
quotient. Moving the x to the denominator worked in the
previous example so let’s try that with this problem as well.
Writing the product in this way gives us a product
that has the form 0/0 in the limit.
So, let’s use L’Hospital’s Rule on the quotient.
Hummmm…. This
doesn’t seem to be getting us anywhere.
With each application of L’Hospital’s Rule we just end up with another
0/0 indeterminate form and in fact the derivatives seem to be getting worse
and worse. Also note that if we
simplified the quotient back into a product we would just end up with either or and so that won’t do us any good.
This does not mean however that the limit can’t be
done. It just means that we moved the
wrong function to the denominator.
Let’s move the exponential function instead.
Note that we used the fact that,
to simplify the quotient up a little. This will help us when it comes time to
take some derivatives. The quotient is
now an indeterminate form of and use L’Hospital’s Rule gives,
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So, when faced with a product we can turn it into a quotient that will allow
us to use L’Hospital’s Rule. However, as
we saw in the last example we need to be careful with how we do that on
occasion. Sometimes we can use either
quotient and in other cases only one will work.