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I am trying to prove the L’Hospitals rule in the following using the mean value theorem of the differential calculus instead of Cauchy's Mean Value Theorem(the generalized mean value theorem of differential calculus).

Consider two functions $f(x)$ and $g(x)$, continuous on a closed interval $[a, b]$ of the X-axis, and differentiable on the interior of that interval. We assume that $g'(x)$ is positive and $f(a) = g(a) = 0$. The ordinary mean value theorem of differential calculus applied separately to $f(x)$ and $g(x)$ furnishes the expression: $$\frac{f(x) - f(a)}{g\left( x \right) - g\left( a \right)}=\frac{f'(c_1)}{g^{'}\left( c_2 \right)}.$$ where $c_1$ and $c_2$ are suitable intermediate values in the open interval $(a, x)$. After taking limit on both sides and substituting $f(a) = g(a) = 0 $, one got $$\lim _{x\rightarrow a+}\frac{f(x) - f(a)}{g\left( x \right) - g\left( a \right)}=\lim _{x\rightarrow a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}$$ However, L’Hospital’s Rule needs $$\lim _{x\rightarrow a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g^{'}\left( x \right)}$$ so I wonder whether $\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g^{'}\left( x \right)}$ hold in this case ?


Some progress I have made in solving the problem :

  1. For every x, there always exist $c_1$ and $c_2$ in (a, x), so one can denote $c_1=m(x)$ and $c_2=n(x)$, then $$\lim _{x\rightarrow a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}=\lim _{x\rightarrow a+}\frac{f'(m(x))}{g^{'}\left( n(x) \right)}$$(As @Bernard pointed out in the comment - if there are several $c_1$s or $c_2$s in $(a, x)$, then I would choose any one of them, then $c_1$ and $c_2$ are functions of $x$ .), so the problem becomes : whether $\lim _{x\rightarrow a+}\frac{f'(m(x))}{g^{'}\left( n(x) \right)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g^{'}\left( x \right)}$ hold in this case ?
  2. As $x$ approaches $a$, $c_1$ and $c_2$ also approach $a$ because it always lies between $a$ and $x$.
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  • $\begingroup$ Writing $\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}$ is meaningless because $c_1$ and $c_2$ are not functions of $x$. $\endgroup$
    – Bernard
    Commented Jan 20, 2019 at 15:20
  • $\begingroup$ @Bernard I think not. For every x, there always exist $c_1$ and $c_2$ in (a, x), so one can denote $c_1=m(x)$ and $c_2=n(x)$. BTW, as $x$ approaches $a$, $c_1$ and $c_2$ also approaches $a$ because it always lies between $a$ and $x$. $\endgroup$
    – iMath
    Commented Jan 20, 2019 at 15:35
  • $\begingroup$ And if there are several $c_1$s or $c_2$s? $\endgroup$
    – Bernard
    Commented Jan 20, 2019 at 15:45
  • $\begingroup$ @Bernard Oh, I haven't considered that. If there are several $c_1$s or $c_2$s, then I may choose any one of them $\endgroup$
    – iMath
    Commented Jan 20, 2019 at 15:53
  • $\begingroup$ With the axiom of choice? It raises some fundamental questions… $\endgroup$
    – Bernard
    Commented Jan 20, 2019 at 16:30

2 Answers 2

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Since we have the same restrictions of domain for $f, g$ we can say that without loss of generality that $c_1 = c_2$.

Fully the proof would be as follows:

If $x \in (a,b)$, then by the Mean Value Theorem applied to our given restrictions of $f,g$ to $[a, x]$, there exists some $\xi_x \in (a,x)$ with $$\frac{f'(\xi_x )}{g'(\xi_x )}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f(x)}{g(x)}$$

Since $\xi_x \rightarrow a$ as $x\rightarrow a_+$ we have $$\lim_{x\rightarrow a_+} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a_+} \frac{f'(\xi_x)}{g'(\xi_x)}$$

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  • $\begingroup$ Sorry, but Cauchy's Mean Value Theorem is the same as the generalized mean value theorem of differential calculus, I don't want to use it to prove here $\endgroup$
    – iMath
    Commented Jan 20, 2019 at 15:50
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You have to use Cauchy's Mean Value Theorem; that is $$\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c)}{g'(c)}$$ for some $c$ such that $a<c<x$. Thus when $x\to a^+$, one has $\to a^+$. So $$\lim _{x\rightarrow a+}\frac{f(x)}{g(x)}=\lim _{x\rightarrow a+}\frac{f'(c)}{g'(c)}=\lim_{c\rightarrow a+}\frac{f'(c)}{g'(c)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g'(x)}.$$

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  • $\begingroup$ Sorry, but Cauchy's Mean Value Theorem is the same as the generalized mean value theorem of differential calculus, I don't want to use it to prove here $\endgroup$
    – iMath
    Commented Jan 20, 2019 at 15:37

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