# L’Hospital rule

I had never heard of L’Hospital rule before I had to teach it, one fateful morning of 2012 at the University of Toronto. The class was your standard Calculus 1 class, where L’Hospital rule features prominently as your one-size-fits-all trick to compute limits. “How come I never heard of this?”, I thought. That’s when I realized that, in more than 10 years of knowing limits, I never felt the need for it: I had Taylor expansions, comparative growths, and so much the better for it. Now, seven years later, I still have not found any relevant use for it. More importantly, after teaching more post-calculus classes, I have grown more and more convinced that putting such emphasis on it is a pedagogical mistake.

## What is it again?

L’Hospital rule allows to compute limits of indeterminate forms of the type $0/0$ or $\infty/\infty$. In a simple form, it reads as follows.

(L’Hospital rule)

If $f$ and $g$ are two differentiable functions defined around a point $x_0$, which can be a real number or $\pm \infty$, and either

$\displaystyle \lim_{x \to x_0} f(x) = \lim_{x \to x_0} g(x) = 0$

or

$\displaystyle \lim_{x \to x_0} f(x) = \lim_{x \to x_0} g(x) = + \infty$,

then, provided the second limit exists,

$\displaystyle \lim_{x \to x_0} \frac{f(x)}{g(x)} = \lim_{x \to x_0} \frac{f'(x)}{g'(x)}$.

It allows to do computations such as the following, where I write “LHR” where L’Hospital is used.

$\displaystyle \lim_{x \to 0} \frac{\sin x}{\log(1+x)} \stackrel{\text{LHR}}{=} \lim_{x \to 0} \frac{\cos x}{1/(1+x)} = \frac{1}{1+0} = 1$

or

$\displaystyle \lim_{x \to + \infty} \frac{e^x}{x^2} \stackrel{\text{LHR}}{=} \lim_{x \to + \infty} \frac{e^x}{2x} \stackrel{\text{LHR}}{=} \lim_{x \to + \infty} \frac{e^x}{2} = + \infty.$

Here is the thing: no mathematician, physicist, economist, or anyone using any type of math, would compute a limit in this way. Actually, they would not compute at all. The first limit is a direct consequence of the fact that $\sin x$ and $\log(1+x)$ look like $x$ when $x$ is small, while the second one is just saying that the exponential grows faster than any power of $x$.

I believe that undergrad calculus courses are so fond of L’Hospital rule for a few reasons.

1. It gives a simple application of derivatives.
2. It allows to compute a bunch of limits.
3. It is an excuse to compute derivatives.

These are very reasonable arguments, but this being said…

1. Aren’t there enough applications of derivatives in a calculus course?
2. Most L’Hospital style limits can be computed in a different and more efficient way. Moreover, are students really excited to compute limits in isolation?
3. Aren’t there enough opportunities to compute derivatives in a calculus course?

## My issues with L’Hospital rule

### Debatable usefulness

I never came across a limit in the wild where I felt the need to use L’Hospital. In every instance, Taylor’s formulas or a bit of thinking give the result more efficiently and eloquently. Let us try an experiment. Open Stewart’s calculus at the section about L’Hospital rule, pick an example at random, and compute the limit with or without L’Hospital. I am ready to bet that 90% of them are more easily obtained without L’Hospital, 10% slightly more easily with L’Hospital, and that these 10% are totally artificial examples.

Let me try: I open Section 4.4 in my 7th edition. Exercises 7-66 are computation of limits, so I take a random number from 7 to 66. I unfortunately happen to get number 29

$\displaystyle \lim_{x \to 0} \frac{\sin^{-1} x}{x}.$

I will come to this one later. I then get number 44,

$\displaystyle \lim_{x \to 0^+} \sin x \ln x.$

It is an indeterminate form of the type $0 \times \infty$. If we wish to apply L’Hospital, we have to turn that into a limit of the type $0 / 0$ or $\infty / \infty$. Which is better? I first tried arbitrarily so write $\sin x \ln x = \sin x / (1/\ln x)$ and after a couple lines, I became really really sad, so let us pretend that we are lucky and do the following computation.

$\displaystyle \lim_{x \to 0^+ } \sin x \ln x = \lim_{x \to 0^+ } \frac{\ln x}{1/ \sin x} \stackrel{\text{LHR}}{=} \lim_{x \to 0^+ } \frac{1/x}{- \cos x/(\sin^2 x)} = - \lim_{x \to 0^+ } \cos x \frac{\sin^2 x}{x}$.

It is a bit better: the $\cos x$ gives us a limit of one, and then L’Hospital again gives

$\displaystyle \lim_{x \to 0^+ } \frac{\sin^2 x}{x} \stackrel{\text{LHR}}{=} \lim_{x \to 0^+ } \frac{2 \sin x \cos x}{1} = \frac{2 \times 0 \times 1}{1} = 0.$

Eventually,

$\displaystyle \lim_{x \to 0^+ } \sin x \ln x = 0$.

Let us summarize what we had to do:

• remember to change $0 \times \infty$ as $0/0$ or $\infty / \infty$;
• be lucky in this endeavor;
• L’Hospital;
• chain rule;
• take away the $\cos x$ since its limit is a nonzero real number;
• L’Hospital again;
• chain rule again.

Another stroke of luck was the fact that the limit is zero, so forgetting a minus sign or a constant would have cost nothing. But more computations means more chances to make a mistake.

Do the computation with L’Hospital after writing $\sin x \ln x = \sin x / (1/\ln x)$.

Let us compare with the following. As $\sin x$ behaves like $x$ as $x \to 0$, then

$\displaystyle \lim_{x \to 0^+} \sin x \ln x = \lim_{x \to 0^+} x \ln x.$

To be totally convincing, one can use equivalents (blog post on this at some point), or merely write $\sin x \ln x = (\sin x/x) \times x \ln x$ and argue that $\sin x / x \to 1$. But now, we only have to compute the limit of $x \ln x$, which we may know. Otherwise, we can use L’Hospital, and need to be careful as before, but it is as least simpler. However, the easiest is to define $y = 1/x$, and conclude

$\displaystyle \lim_{x \to 0^+} x \ln x = \lim_{y \to + \infty} \frac{1}{y} \ln (1/y) = - \lim_{y \to + \infty} \frac{\ln y}{y} = 0$,

where we use that $\ln y$ grows slower than any (positive) power of $y$ as $y \to + \infty$. The amount of computation that we had to do is essentially nonexistent. And we clearly see what is happening: the limit is zero because $\sin x$ behaves likes $x$, and $x$ wins over $\ln x$. Applying L’Hospital hides all of this under a black cloud of symbols.

### Lack of meaning

My second main qualm with L’Hospital rule is that it does not allow one to understand what is going on in a computation. Most problems that appear in math, and even more in applications, are not designed to work smoothly: they are complex and involve abstruse mathematical expressions. Therefore, it becomes necessary to understand what is relevant, and what can be safely ignored. Hence, we have to approximate and compare. For instance, an expression like

$\displaystyle f(x) = \frac{3x^4 - 5x + 28}{x^3 + 7x^2+1}$

will very much resembles $3x$ when $x$ is large, since the highest powers at the numerator and denominator win over the smaller powers. If we want to be more precise, we then compute

$\displaystyle f(x) - 3x = \frac{-21 x^3 + \text{smaller powers}}{x^3 + \text{smaller powers}},$

which implies, for the same reason, that

$\displaystyle \lim_{x \to + \infty} f(x) - 3x = -21.$

We get, without any complex computation, that the graph of $f$ has an oblique asymptote $y = 3x + 21$. Depending on what we wish to do, we may then want to replace the horrible expression or $f$ by $3x$ or $3x + 21$, or even continue and get more and more precise approximations. Applying blindly L’Hospital 3 times to obtain something less precise seem very much suboptimal.

Another example is to approximate $\sin x$ by $x$ if $x$ is small. Remembering that we always use radians, we get

$\displaystyle\sin 10^{\circ} = \sin \frac{10 \pi}{180} \approx \frac{10 \pi}{180} \approx 0.175$

(the real value is $0.174\dots$).

To conclude, thinking in terms of approximations and comparisons not only allows to compute many limits extremely easily, but is also a valuable skill throughout sciences. On the opposite side of the spectrum, L’Hospital rule is just a black box which conceals every relevant information.

### Limited scope

All limits that can be obtained with L’Hospital can be obtained in another way, often more quickly and cleanly. When a limit resists to L’Hospital, it often spells defeat for our students, whereas aforementioned techniques could often nail down the whole problem. The other day, I watched this video on the otherwise excellent YouTube channel Flammable Maths. The host loses his countenance at the sight of the limit

$\displaystyle \lim_{x \to 0^+} \ln^2 x \ln(1+x),$

and eventually computes it with L’Hospital and plenty of trouble. However, we can easily see that this limit is $0$.

Show this in one line.

### So. Many. Mistakes.

My final and maybe most important issue with L’Hospital is that it is seen as a panacea for every single limit, and thought to be simple of use. It is neither, but it is often used without any afterthought, and then disaster strikes. Here are a few factual examples.

#### L’Hospital when it is not an indeterminate form

Consider the following wrong computation.

$\displaystyle \lim_{x \to 3^+} \frac{\ln(x-3)}{\ln(x-2)} = \lim_{x \to 3^+} \frac{1/(x-3)}{1/(x-2)} = \lim_{x \to 3^+} \frac{x-2}{x-3} = + \infty.$   😦

Or we may want to apply L’Hospital again and find

$\displaystyle \lim_{x \to 3^+} \frac{x-2}{x-3} = \lim_{x \to 3^+} \frac{1}{1} = 1$.   😦

The real limit is of the form $- \infty/0^+$, and is therefore $- \infty$.

#### L’Hospital directly on other indeterminate limits

L’Hospital applies only to of the type $0 \times + \infty$. Consider the following wrong computation.

$\displaystyle \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} 1 \times \frac{1}{x} = + \infty$.   😦

As we saw before, the actual limit is 0.

#### L’Hospital to compute derivatives

Consider the following computation.

$\displaystyle \lim_{x \to 0} \frac{\sin^{-1} x}{x} = \lim_{x \to 0} \frac{1/\sqrt{1+x^2}}{1} = 1.$   😦

We should instead recognize that this quantity is simply the derivative of $\sin^{-1}$ at 0, so it is $1/\sqrt{1+0^2} = 1$. It is not a mistake per se, but another instance of “using L’Hospital blindly”.

#### Difficult or absurd computations.

Consider the following computation.

$\displaystyle \lim_{x \to 0} \frac{\sin (x^4)}{(1-\cos x)^2} = \lim_{x \to 0} \frac{4 x^3 \cos(x^4)}{-2(1-\cos x) \sin x} = \cdots$   😦

A Taylor expansion gives directly that this limit is 4. Similarly,

$\displaystyle \lim_{x \to + \infty} \frac{e^x}{x^{10}} = \lim_{x \to + \infty} \frac{e^x}{10x^9} = \lim_{x \to + \infty} \frac{e^x}{90x^8} = \cdots$   😦

As the exponential grows faster than any polynomial, we can tell directly that this limit is $+ \infty$.

#### Instability

Changing an expression in a simple case of L’Hospital can lead to obnoxious difficulties. Computing

$\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{\sin^2 x}$

works well, but a few exponents later and the computation of

$\displaystyle \lim_{x \to 0} \frac{\cos x^3 - 1}{\sin^3 x^2}$

becomes a major pain in the math. However, Taylor expansions readily yields that the limit is $1/2$.

#### L’Hospital does not replace other techniques

Consider the following computation.

$\displaystyle \lim_{x \to + \infty} \frac{\sqrt{x^2+1}}{x} = \lim_{x \to + \infty} \frac{\frac{2x}{2\sqrt{x^2+1}}}{1} = \lim_{x \to + \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x \to + \infty} \frac{1}{\frac{2x}{2\sqrt{x^2+1}}} = \lim_{x \to + \infty} \frac{\sqrt{x^2+1}}{x} = \cdots$   😦

The use of L’Hospital is perfectly justified, but it just does not work. The trick is naturally to simplify the expression to

$\displaystyle \sqrt{1 + \frac{1}{x^2}}$

to see that the limit is $1$. Alternatively, we can say that $x^2+1$ behaves like $x^2$ at $+ \infty$, so the numerator behaves like $x$, and the limit follows.

#### L’Hospital might not work in subtle ways

If we try to apply L’Hospital in the following case, we get

$\displaystyle \lim_{x \to 0} \frac{x^2 \sin(1/x)}{x} = \lim_{x \to 0} - \cos(1/x) + 2 x \sin(1/x)$,   😦

which does not exist. However, by squeeze theorem, we have

$\displaystyle \lim_{x \to 0} \frac{x^2 \sin(1/x)}{x} = \lim_{x \to 0} x \sin(1/x) = 0$.

What is precisely the issue with L’Hospital rule here?

#### Necessity for plenty of tricks

As much as I like to say that L’Hospital rule needs no thinking, it also requires a whole bunch of tricks. There is, as we saw, the problem of turning $0 \times \infty$ into $0/0$ or $\infty/\infty$. One might be easy and the other one impossible, and I do not know of any relevant way to make a choice, except to try and pray. Then there is $\infty - \infty$, which is never easy. A classical example is $\sqrt{x^2+1} - x$, which can be simplified by multiplying by the conjugate quantity to obtain

$\displaystyle \sqrt{x^2+x} - x = \frac{x}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{1+1/x^2}+x} \to \frac12.$

As before, L’Hospital at the third step would lead to total failure. A cleaner way is to factor out an $x$ and use Taylor’s formula on $\sqrt{1+1/x^2}$.

Simplify to find the following limit.

$\displaystyle \lim_{x \to + \infty} (x^3+x^2)^{1/3} - x.$

## An apology (?) for L’Hospital rule

This is a long post, and it is surely controversial. L’Hospital rule is one of the most prized treasures of plenty of undergraduate students, it is simple yet powerful, and it feels very neat when it works. I can think of a few more arguments for when it can come in handy… and of just as many counter arguments.

### It gives classical limits.

Without Taylor expansions, L’Hospital gives classical limits at $0$ such as

$\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{x^2} = - \frac12, \quad \lim_{x \to 0} \frac{e^x-1-x}{x^2} = \frac12, \dots$

I would argue that Taylor’s expansions (or the first terms) should be introduced as soon as possible in any calculus course, and can be given in a formula sheet in an exam.

L’Hospital also gives the comparative growth results, namely, for $\alpha \in (0,+\infty)$,

$\displaystyle \lim_{x \to + \infty} \frac{e^x}{x^{\alpha}} = + \infty, \quad \lim_{x \to + \infty} \frac{\ln x}{x^{\alpha}} = 0, \dots$

On the other hand, these can be obtained from inequalities proven by basic calculus such as $e^x \geq x^n/n!$ for any $n \in \mathbb{N}$ and $x \geq 0.$

### It can be easier for limits at other points

For limits which are not at $0$ or $+ \infty$, L’Hospital does not need a change of variable. For instance

$\displaystyle \lim_{x \to 1} \frac{x^3-1}{x^7-1} \stackrel{\text{LHR}}{=} \lim_{x \to 1} \frac{3x^2}{7x^6} = \frac37$.

This can of course be obtained, with a bit more work, by letting $y = x-1$ and using Taylor’s formula. Moreover, in practice, we very seldom compute limits at other points than $0$ or $+ \infty$. Any case that I can think of (or see in Stewart) when L’Hospital would work faster seems totally artificial.

## Some final questions

What are your own thought about L’Hospital rule? Am I a bit too rough? Why is L’Hospital so popular in some country, but unheard of in others? Is there any natural example where L’Hospital gives a result that could not be obtained more quickly in another way? Let me know!