I Some questions on l'Hôpital rules

[mcastillo356]

TL;DR

Bernouili's work, french mathematician's work, l'Hôpital first rule...Need some more knowledge

Hi, PF

Got questions to start with: ¿some casual background about these Rules?; ¿are them two, as the textbook says?.

https://en.wikipedia.org/wiki/L'Hôpital's_rule (only one statement found)

Here goes the first, from "Calculus, 7th ed, R, Adams, C. Essex"

THEOREM 3 The first l'Hôpital Rule

Suppose the functions $f$ and $g$ are differentiable on the interval $(a,b)$ and $g'(x)\neq 0$ there. Suppose also that
(i) $\lim_{x\to a^+}f(x)=\lim{x\to a^+}g(x)=0$ and
(ii) $\lim_{x\to a^+}{\frac{f′(x)}{g′(x)}}=L$ (where $L$ is finite or $\infty$ or $−\infty$)
Then
$\lim_{x\to a^+}{\frac{f(x)}{g(x)}=L}$
Similar results hold if every occurrence of $\lim_{x\rightarrow {a^+}}$ is replaced by $\lim_{x\rightarrow {b^-}}$ or even $\lim_{x\rightarrow{c^+}}$ where $a<c<b$. The cases $a=-\infty$ and $b=\infty$ are also allowed

PROOF We prove the case involving $\lim_{x\rightarrow{a^+}}$ for finite $a$. Define

$F(x)=\begin{cases}f(x)&\mbox{if}a<x<b\\0&\mbox{if} x=a \end{cases}$

and

$G(x)=\begin{cases}g(x)&\mbox{if}a<x<b\\0&\mbox{if} x=a \end{cases}$

Then $F$ and $G$ are continuous on the interval $[a,x]$ and differentiable on the interval $(a,x)$ for every $x$ in $(a,b)$. By the Generalized Mean-Value Theorem (...) there exists a number $c$ in $(a,x)$ such that
$\frac{f(x)}{g(x)}=\frac{F(x)}{G(x)}=\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F'(c)}{G'(c)}=\frac{f'(c)}{g'(c)}$.

Since $a<c<x$, if $x\rightarrow{a^+}$, then neccesarily $c\rightarrow{a^+}$, so we have

$\lim{x\to{a^+}}{\frac{f(x)}{g(x)}}=\lim{c\to{a^+}}{\frac{f'(c)}{g'(c)}}=L$

Mean Value Theorem seems a limitless tool in Analysis. Question: $\lim{c\to{a^+}}{\frac{f'(c)}{g'(c)}}=L=\frac{f'(c)}{g'(c)}$? Think so. At this point, $c\rightarrow{a^+}$ doesn´t add worth information; it's a useless limit

Attemtp: Wikipedia isn't wrong; is straight, I guess; but incomprehensive for me. I understand what the textbook says, but need some kind of text comment on the aim of my textbook.Thanks. I think LaTeX is not well done, please PF, check it.
Edited at 6:39 AM Europe timing

 

[PeroK]

What's the question?

 

[mcastillo356]

Hi, the questions are: is the limit when $c\rightarrow{a^+}$ of $\dfrac{f'(c)}{g'(c)}$ the same as the quotient $\dfrac{f'(c)}{g'(c)}$ itself? I think so because $a<c$;
Wikipedia truly states only one l'Hôpital Rule? (the article, I mean, the maths at the second paragraph are fine: no need to mention two Rules -fine to me-);
Which source must I pick? I am tempted not to reject anyone; but doubts arise. Let's take 1st l'Hôpital rule: why does look at $x$ when tends to $a$ from the right?; Wikipedia talks about an interval $I$ an $x$ tending, presumably -but not for sure- to $c$ (more intuitive to me).

 

[PeroK]

Hi, the questions are: is the limit when $c\rightarrow{a^+}$ of $\dfrac{f'(c)}{g'(c)}$ the same as the quotient $\dfrac{f'(c)}{g'(c)}$ itself?

Only if the quotient is well defined at $c$. It might be another limit of the form $\frac 0 0$, for example.

 

[Mark44]

Only if the quotient is well defined at $c$. It might be another limit of the form $\frac 0 0$, for example.

Or $\pm \frac \infty \infty$.

 

[mcastillo356]

Only if the quotient is well defined at $c$. It might be another limit of the form $\frac 0 0$, for example.

Could explain some more about this quote? No attempt, clue about it. The question is: how the quotient, and why, should be well defined at ##c#?. What means to be well defined? Don't manage well in Wikipedia.

 

[PeroK]

Could explain some more about this quote? No attempt, clue about it. The question is: how the quotient, and why, should be well defined at ##c#?. What means to be well defined? Don't manage well in Wikipedia.

To take an example: suppose we want to evaluate the following limit using L'Hopital's rule:
$$\lim_{x \to 0} \frac{x^3}{x^2}$$If we differentiate top and bottom, we get another indeterminate form:
$$\lim_{x \to 0} \frac{x^3}{x^2} =\lim_{x \to 0} \frac{3x^2}{2x}$$Your question, if I understand it, is that why don't we drop the limit on the RHS and evaluate the function at $x = 0$? The answer is because we can't as we have another expression of the $\frac 0 0$.

But, we can apply L'Hopital's rule a second time to get:
$$\lim_{x \to 0} \frac{3x^2}{2x} =\lim_{x \to 0} \frac{6x}{2} = 0$$This time we have got a function where the limit can be evaluated simply.

In general, you sometimes have to apply the L'Hopital rule several times.

 

[pasmith]

Since $a<c<x$, if $x\rightarrow{a^+}$, then neccesarily $c\rightarrow{a^+}$, so we have

$\lim{x\to{a^+}}{\frac{f(x)}{g(x)}}=\lim{c\to{a^+}}{\frac{f'(c)}{g'(c)}}=L$

Mean Value Theorem seems a limitless tool in Analysis. Question: $\lim{c\to{a^+}}{\frac{f'(c)}{g'(c)}}=L=\frac{f'(c)}{g'(c)}$? Think so. At this point, $c\rightarrow{a^+}$ doesn´t add worth information; it's a useless limit

But c is not a constant: it is a function of x as it is the result of applying the mean value theorem to a function on [a,x]. So it is more accurate, and perhaps clearer, to write <br /> \lim_{x \to a^{+}} \frac{f(x)}{g(x)} = \lim_{x \to a^{+}} \frac{f&#039;(c(x))}{g&#039;(c(x))}. At this point we can replace c(x) with a dummy variable (either introducing a new symbol or repurposing c or x) to get <br /> \lim_{x \to a^{+}} \frac{f(x)}{g(x)} = \lim_{c \to a^{+}} \frac{f&#039;(c)}{g&#039;(c)}.

 

[mcastillo356]

Your question, if I understand it, is that why don't we drop the limit on the RHS and evaluate the function at $x = 0$? The answer is because we can't as we have another expression of the $\frac 0 0$.

Well, if I've understood, right hand side limit and left hand side limit must be equal to say we have a limit when the argument tends to some $a\in{\mathbb{R}}$. Basically they can differ in many basic limit operations.

 

[mcastillo356]

But c is not a constant: it is a function of x as it is the result of applying the mean value theorem to a function on [a,x]

Brilliant, thanks indeed for the post, I really thought I was facing reals. Nice remark