1
$\begingroup$

Consider the following erroneous usage of L'hopital's rule:

$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{D_h(f(x+h) - f(x))}{D_h(h)} = \lim_{h \to 0} \frac{f'(x+h)}{1} = f'(x) \tag1 $$

Note that if $\lim_{h \to 0} f'(x+h)$ exists, it is always equal to $f'(x)$.

There are two remarks one can make about $(1)$. The usage of L'hopital's rule is entirely fallacious, since we're essentially using a rule which uses the concept of differentiation on the definition of the derivative itself, and that, notwithstanding this, its usage yields the correct answer. Note also that $(1)$ is essentially a generalization of the common (explicit) example of this fallacy where the limit is $\frac{\sin x}{x}$.

My question is twofold:

Is the fact that the circular reasoning above gave us the correct answer entirely a mathematical coincidence, or is there some "deeper" reason behind it? In either case, is there an example in mathematics, analagous to $(1)$, where circular reasoning gives an incorrect answer?

$\endgroup$
4
  • $\begingroup$ The left and right sides are equal by definition. L'Hopital lets you rewrite it and then you rewrite it again. There is no "correct answer" to this equation; there is not a question, in fact. It's debatable whether there is even any circular reasoning used here. $\endgroup$ – Morgan Rodgers Jun 12 '16 at 7:09
  • $\begingroup$ @MorganRodgers I agree. Once you defined $f'(x):=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$ (if exists), then derivatives are defined and L'Hospitals rule can be inferred. After that L'Hospital applies to any $\frac 00$ limit, including this one. $\endgroup$ – gebruiker Jun 12 '16 at 7:13
  • 1
    $\begingroup$ And to answer the question in the title: Circular reasoning always goes wrong. This is not at all suprising, because circular reasoning is the name for a particular kind of logical fallacy. $\endgroup$ – gebruiker Jun 12 '16 at 7:19
  • 1
    $\begingroup$ The answer arrived by circular reasoning will always be correct. By definition circular reasoning is technique where you use the result (say A) which is to be proved as one of the intermediate steps of the proof of A. But the use of circular reasoning is wrong because one does not actually prove result A, but assumes its validity without proof and then thinks that this assumption is equivalent to a proof. $\endgroup$ – Paramanand Singh Jun 12 '16 at 7:27
1
$\begingroup$

Circular reasoning always works. Logic would make for a pretty bad system of deduction if the truth of a proposition $P$ was not a consequence of the hypothesis that $P$ is true!

The notation $P \vdash Q$ means, that from the hypothesis $P$, you can logically deduce $Q$.

$P \vdash P$ is a theorem of logic.


Furthermore, circular reasoning is a good thing. When we learn a subject, such as calculus, starting from first principles we develop and study sophisticated ideas and advanced techniques.

But once we know sophisticated ideas and advanced techniques, they are far easier to use than the basic principles.

e.g. if $P$ a basic fact of calculus (or otherwise something easy to prove at the beginning of your calculus education), it is not uncommon for someone who has learned calculus to realize

$$\text{calculus} \vdash P$$

much more easily than having to work out

$$\text{first principles of calculus} \vdash P$$


The fallacious application of circular logic is when someone tries to make the following argument:

  • $P \vdash P$
  • Therefore, P.

This is an invalid argument form.

$\endgroup$
5
  • 1
    $\begingroup$ I would argue that $P \vdash P$ is not circular reasoning. It only becomes circular when you add: "therefore $P$". $\endgroup$ – gebruiker Jun 12 '16 at 7:34
  • $\begingroup$ Also, I don't see how circular reasoning applies to sturdying calculus. When one starts to study calculus, one assumes that some things are true, without proof (in good faith). The proof will come later in someone's career. I don't think this can be called circular reasoning. $\endgroup$ – gebruiker Jun 12 '16 at 7:40
  • 1
    $\begingroup$ @gebruiker: An example of circular reasoning in the sense I describe is when someone invokes Taylor's series or asymptotic analysis to obtain a more basic result (like a differential approximation or the mean value theorem) $\endgroup$ – user14972 Jun 12 '16 at 7:47
  • $\begingroup$ Ah, I see. That, I can agree with. $\endgroup$ – gebruiker Jun 12 '16 at 7:50
  • 1
    $\begingroup$ I find this answer very strange. You start with "circular reasoning always works", a seemingly provocative statement, but then reveal that you are using a definition of circular reasoning that is not the commonly used definition with which I am familiar (the logical fallacy, which you refer to later). I also don't see how this answer sheds any light on either of the questions asked, so it's strange that the OP accepted this answer. $\endgroup$ – Don Hatch Aug 1 '18 at 3:24
1
$\begingroup$

Circular reasoning is commonly defined to be an argument that assumes that which is to be proven (or supported). The conclusion may be true, false, or indeed meaningless. The invalidity of circular reasoning is independent of whether the conclusion is true or not.

In the immediate Question l'Hôpital's Rule is invoked on the limit of the difference quotient for $f(x)$ that defines $f'(x)$. However nothing explicitly was said to justify this application. While the denominator $h$ is clearly differentiable with respect to $h$ and tends to zero as $h\to 0$, no proof is given that the numerator is differentiable with respect to $h$ and tends to zero as $h\to 0$.

Whether this is an illustration of "circular reasoning" depends on if we are being asked to assume the differentiability of $f$ in order to conclude that property of $f$. The fallacy of such reasoning as presented, with its gaps in justification, is evident from its showing any function $f$ whatever has a derivative.

This of course is one illustration of drawing a false conclusion from circular reasoning, but circular reasoning is invalid even when turned to a purpose of supporting a true proposition. It is not necessary to find a false conclusion in order to disparage an argument for relying on circular reasoning.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.