L'Hôspital's rule – Serlo

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.

L’Hospital's ruleBearbeiten

Theorem (L’Hospital's rule)

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:

  1.   and  
  2.  .

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:

 
To-Do:

theorem fertig schreiben

Examples and applicationsBearbeiten

Standard types   and  Bearbeiten

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:

  1.  
  2.  

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

 

So all in all, there is

 

Exercise (Limits via L’Hospital 3)

Determine the limits

  1.  
  2.  

Solution (Limits via L’Hospital 3)

Part 1:

 

Part 2:

 

Type  Bearbeiten

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.

Exercise (Limits via L’Hospital 4)

Compute

  1.  
  2.  
  3.  

Solution (Limits via L’Hospital 4)

Part 1:

 

Part 2:

 

Part 3:

 

Type  Bearbeiten

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:

 

Exercise (Limits via L’Hospital 5)

Determine

  1.  
  2.  

Solution (Limits via L’Hospital 5)

Part 1: There is

 

Part 2: Here, we have

 

Types  ,  ,   and  Bearbeiten

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:

 

Exercise (Limits via L’Hospital 6)

Compute the following limits

  1.  
  2.  

Solution (Limits via L’Hospital 6)

Part 1:

 

Part 2:

 

Some warningsBearbeiten

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 waysBearbeiten

Growth of exponential and logarithm functionsBearbeiten

Let us consider the limit value

  for   and  

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

  1.  -fold application of the rule of L'Hospital.
  2. 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.

Growth of polynomial functionsBearbeiten

Let us now consider the following limit value of a rational function for  :

 

Here we have, because of   the type  , and by applying the rule of L'Hospital three times we obtain

 

Alternatively, the limit value can be calculated without L'Hospital by excluding and then shortening the highest power ( ):

 

If now in general   and   are normalized polynomials, then there is again

 

If we want to show this with the rule of L'Hospital, we have to apply it a total of   times, and get

 

To calculate the limit without L'Hospital, we can again factor out the highest power, i.e.  , and then calculate the limit value:

 

Exercise (Limits of rational functions)

Show for   that

 

Solution (Limits of rational functions)

Fall 1:  

1st way: without L’Hospital

We factor out the greatest power  , and obtain

 

2nd way: with L’Hospital

We apply the rule of L'Hospital   times, and get

 

Fall 2:  

1st way: without L’Hospital

 

2nd way: with L’Hospital

 

Fall 3:  

1st way: without L’Hospital

We factor out the greatest power  , and obtain

 

2nd way: with L’Hospital

We apply the rule of L'Hospital   times, and get

 

L’Hospital proof might failBearbeiten

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  .

infinite loopsBearbeiten

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

 

L’Hospital makes divergence worseBearbeiten

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!

L’Hospital may render a wrong resultBearbeiten

This can happen whenever the rule is applied although the conditions not are met. An example is

 

Look carefully, in this case   converges towards  , not  ! Since enumerator and denominator are continuous in  , there is

 

L'Hospital is not applicable in the case of   . If you apply the rule anyway, you will get the false result

 

Therefore you should always check first whether the rule of L'Hospital is applicable or whether it is even necessary at all.

Exercise (Limits without 3)

Determine the limits

  1.  
  2.  

What are the limits resulting from the incorrect application of the L'Hospital rule?

Solution (Limits without 3)

Part 1: Inserting results in

 

Since the limit value is of the type  , L'Hospital is not applicable. If we apply the rule anyway, we get the false result

 

Part 2: Since the limit value is of the type  , we first use our standard re-formulation trick

 

For the expression in the exponent there is now  . Because of   the result is

 

In Exponent, L'Hospital is again not applicable. If we apply the rule anyway, we get

 

Because of   the following false result follows

 

Implication: sufficient criterion for differentiabilityBearbeiten

Theorem (Criterion for differentiability)

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  .