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The following is an excerpt from Bartle and Sherbert's Introduction to Real Analysis:

enter image description here

Okay. I am having a little trouble following this proof. What is $c$ and why is it in $(a,b)$? What confuses me is that there is no reference to any $\delta$ nor any reference $x$ such that $0 < x - a < \delta$. They just simply write $L - \epsilon \le \frac{f(\beta)}{g(\beta)} \le L + \epsilon$ and claim that the conclusion follows. I suspect they are using $\beta$ in place of $x$, but this still doesn't account for the absence of a $\delta$ and why $\beta$ wouldn't satisfy $0 < \beta - a < \delta$.

On second thought, perhaps $\beta$ and $c$ are intended to stand for the typically used $x$ and $\delta$, respectively. It seems that what the authors have shown is that for every $\epsilon> 0$, there exists a $c > 0$ such that if $\beta$ satisfies $a < \beta \le c$, then $|\frac{f(\beta)}{g(\beta)} - L| \le \epsilon$. Even on that interpretation, why is $c$ (i.e., $\delta$) in $(a,b)$? In fact, if $a$ and $b$ are negative, this doesn't make much sense...

EDIT:

Okay. I think I follow what you, @Jack, are saying. However, there is one last thing giving me trouble. Just before equation (2), the authors claim that there exists a specific $u \in (\alpha, \beta)$ by the Cauchy Mean Value theorem such that equation (2) holds. Now, in order to validly substitute this in the inequality below (2) to obtain inequality (3), we need that that specific $u$ is in $(a,c)$, but this has not been demonstrated. All that we have is that if $u \in (a,c)$, then $L - \epsilon < \frac{f'(u)}{g'(u)} < L + \epsilon$, but how do we know that the specific $u$, guaranteed to exist by the Cauchy Mean Value theorem, falls in the interval $(a,c)$?

Also, it doesn't seem that we merely want to assume that $\beta \in (a,b)$ but that $\beta$ is arbitrary in $(a,c)$, right? Or may be it doesn't matter, since if we have "$\beta \in (a,b) \implies |\frac{f(\beta)}{g(\beta)} - L | <\epsilon$", then we would clearly have "$\beta \in (a,c) \subseteq (a,b) \implies |\frac{f(\beta)}{g(\beta)} - L | < \epsilon$". Am I thinking about this correctly?

user193319
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2 Answers2

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I recently worked through this exact proof from Bartle's text and had the same confusion as @user193319. In particular, the ordering of the variables was not initially clear. For my own benefit, I rewrote the proof for Case (a) as found below, and found it helpful for my understanding.

Proof: The proof is in four parts as outlined in the figure below. enter image description here

Part 1: From the definition of right-handed limits at a real number and the assumption that $\displaystyle{\!\lim_{x\rightarrow a+}\!f'(x)/g'(x)\!=\!L}$, given any $\varepsilon\!>\!0$, there exists $\delta(\varepsilon/2)\!>\!0$ (forcing $\delta\!<\!(b\!-\!a)$ and letting $d\!\equiv\!a\!+\!\delta$) such that: $$x\in(a,d)\subset(a,b)\ \longrightarrow\ L-\frac{\varepsilon}{2}<\frac{f'(x)}{g'(x)}<L+\frac{\varepsilon}{2}.\qquad(Eq.\ 1)$$

Part 2: Next, consider arbitrary $\alpha,\beta\!\in\!\mathbb{R}$ such that $a\!<\!\alpha\!<\!\beta\!<\!d$. Because $f$ and $g$ are differentiable on $[\alpha,\beta]\!\subset\!I$, they are also continuous on $[\alpha,\beta]$ (Thm. 6.1.2 from Bartle (ed.4)). Therefore, given these $\alpha,\beta$, it follows from the Cauchy Mean Value Thm. (6.3.2 from Bartle) that there exists $c$ in $(\alpha,\beta)$ such that: $$\frac{f'(c)}{g'(c)}=\frac{f(\beta)-f(\alpha)}{g(\beta)-g(\alpha)}.$$ And because $c\!\in\!(\alpha,\beta)\!\subset\!(a,d)$, it follows from Eq. 1's material implication that: $$L-\frac{\varepsilon}{2}<\frac{f(\beta)-f(\alpha)}{g(\beta)-g(\alpha)}<L+\frac{\varepsilon}{2}.\qquad(Eq.\ 2)$$

Part 3: Consider the right-hand limit $\displaystyle{\lim_{\alpha\rightarrow a+}}\!$ $\frac{f(\beta)-f(\alpha)}{g(\beta)-g(\alpha)}$. Given the hypotheses that $\displaystyle{\lim_{x\rightarrow a+}\!\!f(x),g(x)\!=\!0}$, it follows from limit properties (see Thm. 4.2.4 from Bartle, which has an analogous property for one-sided limits) that this limit equals $f(\beta)/g(\beta)$. Therefore, Bartle's Thm. 4.2.6 (which also has an analogous result for one-sided limits) applied to Eq. 2 results in: $$L-\varepsilon<L-\frac{\varepsilon}{2}\leq\frac{f(\beta)}{g(\beta)}\leq L+\frac{\varepsilon}{2}<L+\varepsilon.\qquad(Eq.\ 3)$$

Part 4: To recap, given any $\varepsilon\!>\!0$, there exists $\delta$ that defines $d$ such that $\beta\!\in\!(a,d)$ implies Eq. 3. It directly follows from the definition of right-handed limits that the limit of the quotient $f(x)/g(x)$ as $x$ approaches $a+$ is $L$.

Stephen K.
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    That $d$ and $\epsilon/2$ is not really necessary unless one is a believer in formalism. The proof given there in book is simple enough. – Paramanand Singh Jul 21 '17 at 19:22
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    "Necessary" is a subjective standard defined in the eye of the beholder. I can appreciate that the book's proof is "simple enough", but the "formalism" in this particular case made my understanding of the proof much easier to digest. – Stephen K. Jul 21 '17 at 19:31
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    The only reason you have used $d, \epsilon/2$ is to get strict inequalities with $\epsilon$ instead of the weaker ones the book gets. Right. The distinction between $\leq$ and $<$ in the definition of limit does not matter and both are equivalent. – Paramanand Singh Jul 21 '17 at 19:34
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    Also if this helps you in understanding then you have spent the effort well writing down this proof. +1 – Paramanand Singh Jul 21 '17 at 19:37
  • @ParamanandSingh, indeed your comment on the distinction between $\leq$ and $<$ in the definition of limit is correct. It is curious that almost all texts using the $\varepsilon$-$\delta$ definition of the limit of a function at $a$ use strict inequality ($0!<!|x-a|!<!\delta\rightarrow|f(x)-L|!<!\varepsilon$), whereas it seems simple enough to show that $0!<!|x-a|!<!\delta\rightarrow|f(x)-L|!=!\varepsilon$ is equivalent $\forall\varepsilon!>!0$. – Stephen K. Jul 21 '17 at 19:44
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Well, it's just another way of writing: one could let $\delta=c-a$.

One can write: there exists $\delta>0$ such that for all $x\in(a,b)$ with $0<x-a<\delta$, something is true. Or alternatively, there exists $a<c<b$ such that for all $x\in(a,c)$, something is true.

They are the same: note that in the first case, "$x\in(a,b)$ with $0<x-a<\delta$" is the same as $x\in(a,a+\delta)$ assuming that $\delta>0$ is not very large so that $a+\delta<b$.

[Added for the edited question and request in comment.] I agree that the exposition of the proof is a little bit confusing.

Note that the small argument in the beginning of the proof (before "Case (a)") is true for any $\alpha$ and $\beta$ such that $a<\alpha<\beta<b$. One could take this part out as a lemma. Now choose $c\in(a,b)$ such that
$$ L-\epsilon<\frac{f'(u)}{g'(u)}< L+\epsilon\tag{i} $$ is true for all $u\in(a,c)$. Now by that small lemma, for any $\alpha,\beta$ with $a<\alpha<\beta\leq c$, one has (2) for any $u\in(\alpha,\beta)\subset(a,c)$. (Note that for such $u$ (i) is also true.) Thus one can apply (i) together with (2) to get (3).