Understanding the Proof of L'Hopital's Rule

Solution 1:

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.

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 $L$.

Solution 2:

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).