As a final application of the mean value theorem, more precisely the second mean value theorem, we want to derive L'Hospital's rule. This is a practical way to determine the limit of a quotient by separately differentiating numerator and denominator. The rule is named after the French mathematician Guillaume de l'Hôpital, but was first derived by the Swiss mathematician Johann Bernoulli.
Let and with or with . Let be two differentiable functions where for all . Further, suppose that the limit exists and one of the following two statements is true:
and
.
Then, there is .
Proof (L’Hospital's rule)
We first look at the case with . Because and we can continue the functions and continuously. We obtain the functions with and for all . Further we set and .
We now look at any sequence which converges towards . Since the functions and are continuous, we can apply the generalized mean value theorem. So there is a sequence , such that for all we have or and
Somit folgt
Since this does even hold for every arbitrary sequence , there is in total .
To-Do:
case 2
Now let us consider the case . To do this, we define the auxiliary functions choosing a with for or for . We set or . For all we set and .
In the sequencesden we consider only the case , because the proof for is analogous. There is:
We can therefore apply L'Hospital's rule for the case , which we have already proved. There is:
First, we will deal with the types where the rules can be applied directly.
Example (Limits via L’Hospital 1)
First, some commonly found application: find
There is . In addition, and are differentiable on , and there is . Since further
exists, there is by the theorem von L’Hospital
In general we do not write this down in such detail. We check the requirements in our heads and write down the result as follows:
Hint
With the rule of L'Hospital the limit value can be calculated within "one line". We would like to point out, however, that when applying the rule, we use the derivative
It was therefore precisely this limit value that was needed to calculate it. Since we assume that the derivatives of the basic functions are known once they have been calculated, this is not a problem.
Exercise (Limits via L’Hospital)
Determine the following limits:
Solution (Limits via L’Hospital)
Part 1:
Part 2:
Example (Limits via L’Hospital 2)
Next, we determine
There is . Since also the other conditions for the theorem of L'Hospital are fulfilled, there is
This limit value can be generalized for to
Exercise (Limits via L’Hospital 2)
Determine for the limit
Solution (Limits via L’Hospital 2)
There is
Sometimes it is also necessary to apply the rules of L'Hospital several times in a row before we reach the desired result.
Example (Limits via L’Hospital 3)
Next, we determine
There is . Using L’Hospital, we obtain
Now again, . Using L’Hospital again, we finally obtain
Here, the rules of L'Hospital are not directly applicable. The "trick" is therefore to create a fraction by forming reciprocal values, and thus to obtain a limit value in the standard form or .
Example (Limits via L’Hospital 4)
A standard example is the limit
There is and . We take the reciprocal value
Since there is , we can indeed apply L’Hospital and obtain
Warning
The following term shift
renders an expression . However, by applying the rule of L'Hospital,
This expression is now again of the type , but has a more complicated form than the original one. So the trick does not always lead to success!
Hint
For two arbitrary functions and with respective properties, the re-formulation trick reads:
or
Depending on which of the two forms is used, the limit value can then be calculated more easily.
Next, we will deal with differences of limit values, both of which converge improperly towards . These are often differences of fractional terms. By forming the principal denominator and grouping them into a fractional term, the expression can often be transformed such that the rules of L'Hospital are applicable.
Example (Limits via L’Hospital 5)
Consider the limit
Here we have . So the limit value is of the described type . By forming the principal denominator we obtain
Since we are able to apply L’Hospital and obtain
Hint
For two arbitrary functions and , the re-formulation trick reads:
If one of the cases described occurs, we use the trick we have already used in the calculation of derivative of generalized power functions: We first write as . Since is continuous on all of , the limit can be "pulled inside". The limit formed there is now very often of the kind and can be calculated as described above with the rules of L'Hopital.
Example (Limits via L’Hospital 6)
A common example is the limit
This is a limit of type . As described above we re-formulate it into
If we pull the limit into the exponential function, we get the limit value
This is of type . Above we calculated it as follows
Because of continuity of we hence get
Hint
The limit is the main reason for setting . It makes the function continuous at zero.
Hint
For two arbitrary functions and , the re-formulation trick reads:
However, it is not always useful to apply the rule of L'Hospital. In particular, it should not be applied if the conditions are not met. In this case, the hasty application of the rule may give a false result. We will discuss some warning examples to illustrate this.
L’Hospital proofs can get tedious - there are also other ways
This is of the type and L'Hospital is therefore applicable, which results in
The limit is now again of type . Repeating the rule, we obtain
When applying the rule of L'Hospital, we see the following pattern: the enumerator remains the same except for the pre-factor, but this does not change the divergence behaviour towards . In the denominator the power of decreases by one in every step. If we apply the rule of L'Hospital times, we hence get
But we could have achieved this result much faster and more elegantly. We have already shown above by only one application of L'Hospital
for all . So we get
since .
Hint
The limit value states that every exponential function grows faster than every power function, no matter how large the power is.
Exercise (Limits via L’Hospital 7)
Determine for and the limit
by
-fold application of the rule of L'Hospital.
One-fold application of the rule of L'Hospital and smart re-formulation.
Solution (Limits via L’Hospital 7)
Part 1: The limit value is of type and L'Hospital is applicable. There is
The limit value is again of the type , the power in the numerator decreases by one, that in the denominator remains the same. After -fold application of L'Hospital we obtain
Part 2: There is
for all . Hence
As .
Hint
The limit value says that the logarithmic function grows slower than any power function, no matter how small the power is.
In this section we will present some examples of limit values where the rule of L'Hospital "fails". This can happen because the rule of L'Hospital is a sufficient but not a necessary condition for the existence of the limit value .
Sometimes the rule of L'Hospital can go in an "infinite loop". An example is
This limit is of the type , and L'Hospital is applicable. If we do so, we will obtain
The resulting limit value is now again of the type . If we look more closely, we see that the enumerator and denominator have been changed by the use of L'Hospital. If we now apply the rule again, the result is
The original limit value is therefore not changed. The rule of L'Hospital therefore does not help us with this limit value! The reason is that exponentials do not change under differentiation. However, there is a relatively simple way to reach the destination without L'Hospital:
If we factor out from the numerator and denominator, and then shorten it, we get
Exercise (Limits without L’Hospital 1)
Determine the limit
What is the problem in applying the rule of L'Hospital?
Solution (Limits without L’Hospital 1)
By factoring out in the denominator we obtain
The application of the rule of L'Hospital leads into an endless loop, because
It may also happen that the use of L'Hospital actually "makes the situation worse". In other words, a limit value that exists can be transformed by applying the rule into a limit value that no longer exists. Therefore always note: From we get , but not vice versa. In particular, the fact that does not exist, does not imply that does not exist. Let us look at
There is . Therefore we have a type . Application of L'Hospital now renders
Now we have a problem because this limit value does not exist. Let us look at the sequences with
This certainly diverges to . However, there is
This limit value does not exist (not even improperly), since diverges. This means that L'Hospital is inapplicable here too. That the original limit value does exist can be seen from the following conversion trick: Because of there is
Exercise (Limits without L’Hospital 2)
Determine
What is the problem in applying the rule of L'Hospital?
Solution (Limits without L’Hospital 2)
First there is with a simple re-formulation trick
Further, on the one hand , which follows directly from the estimation . On the other hand , which we have shown with the rule of L'Hospital above.
In total we get
If we apply the rule of L'Hospital, which is allowed since the limit value is of the type , we obtain
This limit now diverges because the () expression for diverges. Namely, for the null sequence there is . So the application of L'Hospital was again unsuccessful!
Let be an open interval and . Further let be continuous in and differentiable in . In addition, let there be . Then f is differentiable also at and there is .
Proof (Criterion for differentiability)
We have to show:
Now is and . Furthermore, and are differentiable for , and . With the rule of L'Hospital we obtain
Thus is differentiable at with .
Alternative proof (Criterion for differentiability)
We can also use the mean value theorem to show
Let for this be a sequence in with . Then according to the precondition for all is continuous on (or ) and differentiable on (or ). According to the mean value theorem there is for all a (or ) with
Since now and (or ), there is also . Because of it follows that . So we get
Since with was arbitrary, we obtain
Hint
A function that fulfils the criterion from the theorem is not only differentiable at the point . Because of the derivative function is even continuous in . Therefore the criterion is sufficient and not necessary for the differentiability in .
Question: Give an example of a differentiable function that does not satisfy the conditions of the theorem.
We are looking for a differentiable function whose derivative is not continuous at one point. An example (among many) is the function
It has at the derivative , but is not continuous there, as the limit
does not exist.
Exercise (Differentiability of the Si-function)
Let
Show, without using the differential quotient, that is differentiable at zero and calculate the derivative .
Solution (Differentiability of the Si-function)
Step 1: is continuous at zero
By L’Hospital there is
So is continuous at zero.
Step 2: is differentiable on
Since and are differentiable on , the function is also differentiable there, by the quotient rule. Further there is for :
Step 3: is differentiable at zero
We use the criterion from the previous theorem. (Because of Step 1 and 2, it is applicable.) There is
According to the criterion, is differentiable at zero with .