Serlo: EN: L'Hôspital's rule

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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.

Examples and applications

Standard types $\frac{0}{0}$ and $\frac{\pm \mathrm{\infty }}{\pm \mathrm{\infty }}$

First, we will deal with the types where the rules can be applied directly.

Sometimes it is also necessary to apply the rules of L'Hospital several times in a row before we reach the desired result.

Type $0\cdot (\pm \mathrm{\infty })$

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 $\frac{0}{0}$ or $\frac{\pm \mathrm{\infty }}{\pm \mathrm{\infty }}$.

Type $\mathrm{\infty }-\mathrm{\infty }$

Next, we will deal with differences of limit values, both of which converge improperly towards $\mathrm{\infty }$. 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.

Types $0^0$, $0^{\mathrm{\infty }}$, ${\mathrm{\infty }}^0$ and $1^{\mathrm{\infty }}$

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 $f^g$ as $\exp \circ (g\cdot \ln \circ f)$. Since $\exp$ is continuous on all of $ℝ$, the limit can be "pulled inside". The limit formed there is now very often of the kind $0\cdot \pm \mathrm{\infty }$ and can be calculated as described above with the rules of L'Hopital.

Some warnings

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

Growth of exponential and logarithm functions

Let us consider the limit value

This is of the type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ and L'Hospital is therefore applicable, which results in

The limit is now again of type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$. 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 $\mathrm{\infty }$. In the denominator the power of $x$ decreases by one in every step. If we apply the rule of L'Hospital $k$ 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 $\alpha >0$. So we get

since $\tilde{\alpha }=\frac{\alpha }{k}>0$.

Growth of polynomial functions

Let us now consider the following limit value of a rational function for $x\to \mathrm{\infty }$:

Here we have, because of $\underset{x\to \mathrm{\infty }}{\lim }{x}^3-4x+1=\mathrm{\infty }=\underset{x\to \mathrm{\infty }}{\lim }{x}^3-3x^2+2x-4$ the type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, 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 ($x^3$):

If now in general $p(x)=x^n+a_{n-1}{x}^{n-1}+\dots +a_1{x}^1+a_0$ and $q(x)=x^n+b_{n-1}{x}^{n-1}+\dots +b_1{x}^1+b_0$ 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 $n$ times, and get

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

L’Hospital proof might fail

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 $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$.

infinite loops

Sometimes the rule of L'Hospital can go in an "infinite loop". An example is

This limit is of the type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, and L'Hospital is applicable. If we do so, we will obtain

The resulting limit value is now again of the type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$. 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 $e^x$ from the numerator and denominator, and then shorten it, we get

L’Hospital makes divergence worse

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 $\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}=c$ we get $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=c$, but not vice versa. In particular, the fact that $\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$ does not exist, does not imply that $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$ does not exist. Let us look at

There is $2x\pm \sin (x)\ge 2x-1\to \mathrm{\infty }$. Therefore we have a type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ . Application of L'Hospital now renders

Now we have a problem because this limit value does not exist. Let us look at the sequences $(x_n{)}_{n\in \mathbb{N}}$ with

This certainly diverges to $\mathrm{\infty }$. However, there is

This limit value does not exist (not even improperly), since $(-1{)}^{n+1}$ 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 $\underset{x\to \mathrm{\infty }}{\lim }\frac{\sin (x)}{x}=0$ there is

L’Hospital may render a wrong result

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

Look carefully, in this case $x$ converges towards $\pi$, not $0$! Since enumerator and denominator are continuous in $\pi$, there is

L'Hospital is not applicable in the case of $\frac{0}{\pi }$ . 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.

Implication: sufficient criterion for differentiability

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