Some questions on l'Hôpital rules

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mcastillo356
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TL;DR Summary
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
 
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Answers and Replies

  • #2
PeroK
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What's the question?
 
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  • #3
mcastillo356
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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).
 
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PeroK
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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.
 
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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##.
 
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mcastillo356
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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.
 
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PeroK
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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.
 
  • #8
pasmith
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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 [itex]c[/itex] is not a constant: it is a function of [itex]x[/itex] as it is the result of applying the mean value theorem to a function on [itex][a,x][/itex]. So it is more accurate, and perhaps clearer, to write [tex]
\lim_{x \to a^{+}} \frac{f(x)}{g(x)} = \lim_{x \to a^{+}} \frac{f'(c(x))}{g'(c(x))}.[/tex] At this point we can replace [itex]c(x)[/itex] with a dummy variable (either introducing a new symbol or repurposing [itex]c[/itex] or [itex]x[/itex]) to get [tex]
\lim_{x \to a^{+}} \frac{f(x)}{g(x)} = \lim_{c \to a^{+}} \frac{f'(c)}{g'(c)}.[/tex]
 
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  • #9
mcastillo356
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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.
 
  • #10
mcastillo356
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But [itex]c[/itex] is not a constant: it is a function of [itex]x[/itex] as it is the result of applying the mean value theorem to a function on [itex][a,x][/itex]
Brilliant, thanks indeed for the post, I really thought I was facing reals. Nice remark
 

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