# Avoiding circular logic using L'Hospital's rule

Often, using L'Hospital's rule can make a limit much simpler to evaluate, but in some circumstances it can be incorrect to use the rule even when all of its criteria are met - one example being the evaluation of $$\lim_{x\to0}\frac{\sin x}{x}$$, which relies on $$\frac{d}{dx}\sin x=\cos x$$ being known, which itself relies on the limit we are trying to prove!

How, when considering using L'Hospital's rule, can examples like this one be spotted in order to avoid circular logic?

At first glance, I wouldn't have known that the proof was circular, and I'm concerned that I might make similar mistakes with other functions.

• Do you mean $\lim _{x\to 0}\frac{\sin x}{x}$? Because $\lim_{x\to 0}\sin x$ shouldn't require anything more than a knowledge of continuity of $\sin$ to evaluate. Mar 7, 2017 at 17:30
• Generally, I think that L'Hoptial's rule is a blunt tool, and other methods to find a limit ultimately show more insight, and are preferable. Sometimes, it is the cleanest and the obvious choice. As far as circular logic goes, I think it depends on the context and the audience. If someone has asked to evaluate $\lim_\limits{x \to 0} \frac {sin x}{x}$ Then it seems obvious that the do not already know that $\frac {d}{dx} sin x = cos x$ or they wouldn't be asking the question. Mar 7, 2017 at 17:34
• @DougM: I wouldn't necessarily conclude that they don't know the derivative of $\sin x$. Sometimes they just don't recognize the definition of the derivative, whereas if it were written as $\lim_{x \to 0} \frac{\sin x - \sin 0}{x - 0}$, they would see it. The insight that $\sin 0 = 0$ and that that can be used to put the limit in the form of the derivative—that's what's required. Mar 7, 2017 at 17:47
• @acernine: I think it depends on whether you just want the answer, or you want to demonstrate/understand the answer. Just because a particular application of L'Hopital's is circular doesn't mean the answer is incorrect. Mar 7, 2017 at 17:49
• @DougM: But why do most students know that $\frac{d}{dx} \sin x = \cos x$? Because it's in a table they've memorized (or tried to). Maybe at some point, they were shown the derivation, but if so, the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ was quickly used and probably just as quickly forgotten. The fact that it was required doesn't mean that they remember it, or that they will recognize it. Mar 7, 2017 at 18:07

Use of the rule can never be incorrect in the situations you're describing.

One may argue that in some cases doesn't tell you anything you didn't already know, but that does not mean that the conclusions you reach from it are in any danger of being false.

At worst you can say that using L'Hospital's rule on $\lim_{x\to 0} \frac{\sin x}{x}$ is a detour compared to recognizing the original limit as being the definition of $\sin'(x)$ at $0$ -- but that doesn't put the truth of the result at risk. The limit IS the derivative of the sine, no matter whether you reach this conclusion by L'Hospital or by pattern-matching the definition of a derivative.

If you have a valid way of finding the derivative other than applying the definition directly (and this will usually be the case; it is extremely rare to need to calculate derivatives from first principles rather than symbolically), then it doesn't matter how you discovered that this derivative is what you're looking for.

Of course if what you're doing is learning for the first time what the derivative of the sine function is, then L'Hospital will not help you. This is not because it is not valid, but because what you can conclude then is at most that what you're looking for is the thing you were looking for -- which is true but useless. (And even that depends on actually knowing that $\sin$ is continuously differentiable in the fist place).

If at that point, in that specific context you decide to proceed by "and we know the derivative of sine is cosine", then you will be guilty of circular reasoning. But the error is then not that you used L'Hospital, but that external to your use of L'Hospital you decided to assume something you hadn't actually finished proving yet.

This is a slightly different take on this question than the other nicely written answer by Henning Makholm.

If you are trying to evaluate any limit of the form $$\lim_{x\to a} \frac{f(x) - f(a)} {x-a}$$ via L'Hospital's Rule then you are doing things in a pretty roundabout way. This is more like getting hold of your right ear with your left hand and that too by going through the back of your head. It is preferable to use your right hand to get hold of your right ear. Usage of L'Hospital's Rule in this case is correct only when $f'$ is continuous at $a$ and it will fail when this is not the case (try with $f(x) =x^{2}\sin(1/x),f(0)=0,a=0$).

Some people prefer to say that this kind of usage of L'Hospital's Rule is circular. This is not exactly the case because the limit in question is $f'(a)$ and L'Hospital's Rule never uses the value $f'(a)$, but rather it uses values of $f'(x)$ for $x\neq a$ and then we take limit of $f'(x)$ as $x\to a$. In most common cases the value $f'(x)$ is obtained by a formula which is also applicable when $x=a$ and hence it just makes more sense to use that formula to calculate $f'(a)$ rather than evaluating the limit of $f'(x)$ as $x\to a$.

For some specific functions $f$ it may be possible that the formula for $f'(x)$ for general $x$ is dependent on the evaluation of $f'(a)$ and in these cases the use of L'Hospital's Rule is definitely circular. Thus if the function $f(x) = \sin x$ is defined using geometrical definition then the evaluation of the derivative of $\sin x$ is crucially dependent on the limit formula $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ and then the usage of L'Hospital's Rule for evaluation of this limit is indeed circular.

• The other nicely written answer. L'Hôpital's rule does not require continuity of the derivatives; the problem in your case is that the limit doesn't exist, not that the derivative is discontinuous. Aug 16, 2023 at 22:33
• @LSpice: well if the limit of $f'(x)$ exists and equals $L$ then by mean value theorem one automatically has $f'(a) =L$ so that the derivative is continuous if its limit exists. So in order to show the problem I must take an example where limit of derivative doesn't exist. Aug 17, 2023 at 0:20
• Re, if $\lim_{x \to a} f'(x)$ exists and $f'(a)$ is defined, then $f'$ is continuous at $x = a$; but $\lim_{x \to a} f'(x)$ can exist without $f'(a)$ being defined. Aug 17, 2023 at 2:03
• @LSpice: that happens only when $f$ is discontinuous at $a$ and then we know beforehand that the derivative $f'(a)$ does not exist. More precisely if $f$ is continuous in a neighborhood of $a$ and $\lim_{x\to a} f'(x) =L$ then $f'(a)$ exists and equals $L$. Aug 17, 2023 at 4:13