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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:
$$frac{f(x) - f(a)}{gleft( x right) - gleft( a right)}=frac{f'(c_1)}{g^{'}left( c_2 right)}.$$
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
$$lim _{xrightarrow a+}frac{f(x) - f(a)}{gleft( x right) - gleft( a right)}=lim _{xrightarrow a+}frac{f(x)}{gleft( x right)}=lim _{xrightarrow a+}frac{f'(c_1)}{g^{'}left( c_2 right)}$$
However, L’Hospital’s Rule needs $$lim _{xrightarrow a+}frac{f(x)}{gleft( x right)}=lim _{xrightarrow a+}frac{f'(x)}{g^{'}left( x right)}$$
so I wonder whether $lim _{xrightarrow a+}frac{f'(c_1)}{g^{'}left( c_2 right)}=lim _{xrightarrow a+}frac{f'(x)}{g^{'}left( x right)}$ 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 $$lim _{xrightarrow
   a+}frac{f(x)}{gleft( x right)}=lim _{xrightarrow
   a+}frac{f'(c_1)}{g^{'}left( c_2 right)}=lim _{xrightarrow
   a+}frac{f'(m(x))}{g^{'}left( n(x) right)}$$(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 $lim _{xrightarrow
   a+}frac{f'(m(x))}{g^{'}left( n(x) right)}=lim _{xrightarrow
   a+}frac{f'(x)}{g^{'}left( x right)}$ hold in this case ?


2. As $x$ approaches $a$, $c_1$ and $c_2$ also approach $a$ because it
   always lies between $a$ and $x$.

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 $xrightarrow a_+$ we have $$lim_{xrightarrow a_+} frac{f(x)}{g(x)}=lim_{xrightarrow a_+} frac{f'(xi_x)}{g'(xi_x)}$$

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 $xto a^+$, one has $to a^+$. So
$$lim _{xrightarrow a+}frac{f(x)}{g(x)}=lim _{xrightarrow a+}frac{f'(c)}{g'(c)}=lim_{crightarrow a+}frac{f'(c)}{g'(c)}=lim _{xrightarrow a+}frac{f'(x)}{g'(x)}.$$