# About Theorem 5.13 in “Principles of Mathematical Analysis” by Walter Rudin L'Hospital's Rule L'Hopital's Rule

I am reading "Principles of Mathematical Analysis" by Walter Rudin.

Thank you Saaqib Mahmood.
I copied and pasted your text

Theorem 5.13 on p.109:

Suppose $$f$$ and $$g$$ are real and differentiable in $$(a, b)$$, and $$g^\prime(x) \neq 0$$ for all $$x \in (a, b)$$, where $$-\infty \leq a < b \leq +\infty$$. Suppose $$\frac{f^\prime(x)}{g^\prime(x)} \to A \ \mbox{ as } \ x \to a. \tag{13}$$ If $$f(x) \to 0 \ \mbox{ and } \ g(x) \to 0 \ \mbox{ as } \ x \to a, \tag{14}$$ or if $$g(x) \to +\infty \ \mbox{ as } \ x \to a, \tag{15}$$ then $$\frac{f(x)}{g(x)} \to A \ \mbox{ as } \ x \to a. \tag{16}$$ The analogous statement is of course also true if $$x \to b$$, or if $$g(x) \to -\infty$$ in (15). Let us note that we now use the limit concept in the extended sense of Definition 4.33.

Here is Definition 4.33:

Let $$f$$ be a real function defined on $$E \subset \mathbb{R}$$. We say that $$f(t) \to A \ \mbox{ as } \ t \to x,$$ where $$A$$ and $$x$$ are in the extended real number system, if for every neighborhood $$U$$ of $$A$$ there is a neighborhood $$V$$ of $$x$$ such that $$V \cap E$$ is not empty, and such that $$f(t) \in U$$ for all $$t \in V \cap E$$, $$t \neq x$$.

And, here is Rudin's proof:

We first consider the case in which $$-\infty \leq A < +\infty$$. Choose a real number $$q$$ such that $$A < q$$, and then choose $$r$$ such that $$A < r < q$$. By (13) there is a point $$c \in (a, b)$$ such that $$a < x < c$$ implies $$\frac{ f^\prime(x) }{ g^\prime(x) } < r. \tag{17}$$ If $$a < x < y < c$$, then Theorem 5.9 shows that there is a point $$t \in (x, y)$$ such that $$\frac{ f(x)-f(y) }{ g(x)-g(y) } = \frac{f^\prime(t)}{g^\prime(t)} < r. \tag{18}$$ Suppose (14) holds. Letting $$x \to a$$ in (18), we see that $$\frac{f(y)}{g(y)} \leq r < q \qquad \qquad \qquad (a < y < c) \tag{19}$$

Next, suppose (15) holds. Keeping $$y$$ fixed in (18), we can choos a point $$c_1 \in (a, y)$$ such that $$g(x) > g(y)$$ and $$g(x) > 0$$ if $$a < x < c_1$$. Multiplying (18) by $$\left[ g(x)- g(y) \right]/g(x)$$, we obtain $$\frac{ f(x) }{ g(x) } < r - r \frac{ g(y) }{g(x)} + \frac{f(y)}{g(x)} \qquad \qquad \qquad (a < x < c_1). \tag{20}$$ If we let $$x \to a$$ in (20), (15) shows that there is a point $$c_2 \in \left( a, c_1 \right)$$ such that $$\frac{ f(x) }{ g(x) } < q \qquad \qquad \qquad (a < x < c_2 ). \tag{21}$$

Summing up, (19) and (21) show that for any $$q$$, subject only to the condition $$A < q$$, there is a point $$c_2$$ such that $$f(x)/g(x) < q$$ if $$a < x < c_2$$.

In the same manner, if $$-\infty < A \leq +\infty$$, and $$p$$ is chosen so that $$p < A$$, we can find a point $$c_3$$ such that $$p < \frac{ f(x) }{ g(x) } \qquad \qquad \qquad ( a< x < c_3), \tag{22}$$ and (16) follows from these two statements.

Rudin didn't write $$g(x) - g(y) \neq 0$$ for any $$x, y$$ such that $$a < x < y < b$$ in $$\frac{ f(x)-f(y) }{ g(x)-g(y) } = \frac{f^\prime(t)}{g^\prime(t)} < r. \tag{18}$$

Is this fact so obvious?
If so, please tell me the reason why this fact is so obvious.

I didn't think this fact was so obvious, so I proved:

By assumption, $$g'(x) \neq 0$$ on $$(a, b)$$.
Let $$x, y$$ be any real number such that $$a < x < y < b$$.
Let $$x', y'$$ be any real number such that $$a < x' < x$$ and $$y < y' < b$$.
Then, $$g$$ is a differentiable function on $$[x', y']$$.
By the Intermediate Value Theorem for derivatives (Theorem 5.12 on p.108),
$$g'(x) > 0$$ for all $$x \in [x', y']$$ or $$g'(x) < 0$$ for all $$x \in [x', y']$$.
So, $$g$$ is strictly monotonically increasing on $$[x', y']$$ or $$g$$ is strictly monotonically decreasing on $$[x', y']$$.
So $$g(x) < g(y)$$ or $$g(x) > g(y)$$.
So, $$g(x) - g(y) \neq 0$$.

• By the Mean Value Theorem, it's obvious. As if $g(x)=g(y)$, $g'(s)=0$ for $s \in (x,y)$ – YuiTo Cheng Feb 17 at 3:20
• @YuiToCheng Thank you very much for your answer. – tchappy ha Feb 17 at 3:31
• Try to use the Mean Value Theorem more when you are stuck. – YuiTo Cheng Feb 17 at 3:40
• This more general form of L'Hospital's rule is often useful, e.g. see here and here and here. – Bill Dubuque Feb 17 at 4:16