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So we all know that, if $f$ and $g$ are defined on $[a,b]$ and are differentiable at a point $x \in [a,b]$, then $f/g$ is differentiable at $x$ if $g(x) \neq 0$. And the quotient rule says that $\left (\frac{f}{g} \right)'(x) = \frac{g(x) f'(x) - g'(x) f(x)}{g^2(x)}$.

Rudin's proof of Theorem 5.3 on page 105 says that, let $h = f/g$, then \begin{equation} \lim_{t \to x} \frac{h(t)-h(x)}{t-x} = \lim_{t \to x} \frac{1}{g(t)g(x)} \left[ g(x) \frac{f(t)-f(x)}{t-x} - f(x) \frac{g(t)-g(x)}{t-x} \right] = \frac{g(x) f'(x) - g'(x) f(x)}{g^2(x)}. \end{equation}

I understand why the algebra here is correct, but I can't see why my work below is wrong (although it certainly is). I have \begin{equation} \varphi(t) = \frac{\frac{f(t)}{g(t)}-\frac{f(x)}{g(x)}}{t-x} = \frac{f(t)}{g(t) \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)}. \end{equation} Then, \begin{align} \lim_{t \to x}\varphi(t) & = \lim_{t \to x} \left[ \frac{f(t)}{g(t) \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)} \right]\\ & = \lim_{t \to x} \frac{f(t)}{g(t) \cdot (t-x)} - \lim_{t \to x} \frac{f(x)}{g(x) \cdot (t-x)}\\ & = \lim_{t \to x} \frac{1}{g(t)} \cdot \lim_{t \to x} \frac{f(t)}{t-x} - \lim_{t \to x} \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(x)}{t-x}\\ & = \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(t)}{t-x} - \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(x)}{t-x}\\ & = \frac{1}{g(x)} \cdot \left( \lim_{t \to x} \frac{f(t)}{t-x} - \lim_{t \to x} \frac{f(x)}{t-x} \right)\\ & = \frac{1}{g(x)} \cdot \left( \lim_{t \to x} \frac{f(t) - f(x)}{t-x} \right)\\ & = \frac{f'(x)}{g(x)}. \end{align}

I believe there must be some algebraic errors during the computation, but I can't see where. Thanks for any help!

TonyK
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Robert
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    In general, do not use any rule/theorem without verifying the hypothesis. – Kavi Rama Murthy Apr 04 '25 at 23:44
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    it is great that you try to prove this rule by your own method and you will benefit from analyzing the errors – miracle173 Apr 05 '25 at 08:55
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking / debugging machine. – Bill Dubuque Apr 06 '25 at 20:34
  • For the record, I voted to close it an an SV missing context, not as a duplicate. $\ \ $ – Bill Dubuque Apr 06 '25 at 21:37
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    Voting to reopen. This is an interesting question. – TonyK Apr 06 '25 at 21:40
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    @BillDubuque How is this an SV question? It doesn't even pass the "proposed solution" screen of that tag. The motivation of the question seems clear enough: a math student is noodling around with different approaches to already-solved problems in order to develop proof skills. This derivation doesn't work and they know it, but in order to learn from their mistake they first have to have a clue where the mistake is. – David K Apr 06 '25 at 21:59
  • @DavidK Sure it does. Whether or not the OP thinks the proof is correct or not makes no difference - it still falls under the SV policy. – Bill Dubuque Apr 06 '25 at 22:06

3 Answers3

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Disregarding all the rearranging that's going on, your argument boils down to $$\lim_{t \to x} \left[ \frac{f(t)}{g(t) \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)} \right]\\ =\lim_{t \to x} \left[ \frac{f(t)}{\left[\lim_{t\to x}g(t)\right] \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)} \right]$$ which is an illegal operation--you can't first apply the limit to part of your expression, and then apply the limit to what remains.

A similar erroneous move would allow you to argue: $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left(1+\lim_{n\to\infty}\frac1n\right)^n=\lim_{n\to\infty}1^n=\lim_{n\to\infty}1=1.$$

grand_chat
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  • Please strive not to answer off-topic solution verification questions (doing so violates site policy). – Bill Dubuque Apr 06 '25 at 20:23
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    For heaven's sake @Bill! The OP had a genuine problem, well stated, and grand_chat helped them out with it. What is it with you? – TonyK Apr 06 '25 at 20:51
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    @TonyK There is nothing "wrong" with moderators who strive to keep the site healthy by informing users about site policies. – Bill Dubuque Apr 06 '25 at 21:59
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    If questions of this kind are considered as off-topic and will be discouraged, then there's something wrong with the policy of this site, which was built for helping people when they have problems with math. – Yang Xiaosheng Apr 07 '25 at 09:37
  • @Yang This site is designed as a Q&A site - not a broad spectrum tutoring site. See the meta site for much discussion on such matters, including the compromises made on related policy matters. – Bill Dubuque Apr 07 '25 at 17:17
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    @BillDubuque I don't see why this question is not a typical question that should be refused by a Q&A site. And why on earth does a Q&A site refuse certain kinds of answers? Just look at how many people upvotes this question and the following answers!! Doesn't this mean that this post has helped many people? Just imagine if this post is deleted, where should we find so many helpful and generous people and helpful answers?? Again, if what you says aligns the site policy completely, then there's something wrong with the site policy. – Yang Xiaosheng Apr 08 '25 at 05:15
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You are splitting the limit of the difference as the difference of the limits. But the theorem you need to do that says the following:

$$\lim_{x\to a} (f(x)-g(x)) = \lim_{x\to a} f(x)-\lim_{x\to a} g(x) $$

as long as the limits $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ exist.

Later you have the same problem with the product of limits.

Matthew Towers
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jjagmath
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    To make things a bit more explicit, the OP's proof involves the quantities $\lim_{t \to x} \frac{f(t)}{t-x}$ and $\lim_{t \to x} \frac{f(x)}{t-x}$; but it's fairly clear that these limits do not exist in the case where $f(x) \neq 0$. – Michael Seifert Apr 05 '25 at 12:47
  • Please strive not to answer off-topic solution verification questions (doing so violates site policy). – Bill Dubuque Apr 06 '25 at 20:23
  • @BillDubuque User is not asking about a solution verification. – jjagmath Apr 06 '25 at 20:36
  • @jjagmath Sure it is - debugging is part of verification, and the same context requirements apply. – Bill Dubuque Apr 06 '25 at 21:23
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    @BillDubuque No, it's not. The OP is not posting a proof to verify. OP is asking to find an error. "solution-verification" tag says that OP is expected to specify which part is in doubt. In this post OP is aware that the proof is incorrect and asking to pinpoint the error. Is the OP supposed to specify which part is wrong? If OP knew which part is wrong won't be asking in the first place. – jjagmath Apr 06 '25 at 21:34
  • The point of the policy is that questions asking to check/debug a proof are not a good fit for this Q&A site. It doesn't matter whether or not you know the proof is correct or not. See the meta site for further discussion. – Bill Dubuque Apr 06 '25 at 21:40
  • @BillDubuque I see this recent discussion. Did you have something else in mind? – David K Apr 06 '25 at 22:35
  • @David No, not that post. There have been prior meta discussions explaining why such context is required for proof check/debug questions. Here, at the least, instead of saying "I can't see why it is wrong" the OP should have stated their justifications for all the steps in the proof. Often doing so the OP will find the error before posting, or at least narrow it down to the step that is in doubt. Further, such context likely provides enough info for answerers to accurately pinpoint the source of the problem (often there are wrong guesses when such context is lacking). – Bill Dubuque Apr 07 '25 at 01:40
  • @BillDubuque If you think the question does not follow the policy of the forum, how is that you haven't vote to close it? – jjagmath Apr 07 '25 at 02:31
  • @jjagmath I did vote to close (as an SV lacking context) - see the timeline. – Bill Dubuque Apr 07 '25 at 02:33
  • @BillDubuque Those are good points. It is certainly a good policy to ask people to show more of their work in a question when steps are not fully justified, as in this case, whether or not we consider the question SV. I've done that elsewhere; I apologize for neglecting it here. It is my guess that the answer we are commenting on pinpoints the detail that OP overlooked--but without seeing OP's complete thoughts, I admit I can only guess. (OP suggested an algebra error; I don't think that's what happened.) – David K Apr 07 '25 at 02:51
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In general, it is good practice to instantiate your proof on concrete examples. Here, say both $f$ and $g$ are the identity.

You can then for instance see that your first line evaluates to $0$, while the second gives you $+ \infty - \infty$.

And congrats, trying things by yourself and asking for feedback on your proofs is great practice :-)

Corto
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