Taking Limits of Expressions of Different Variables in Proof of L'Hopital's Rule? In my real analysis textbook, they present the proof of L'Hopital's by using Cauchy's mean value theorem. The part in the proof I am struggling to work out rigorously is a particular implication involving equivalent limits using different variables.
We have assumed that $f(c)=g(c)=0$, $f$ and $g$ are differentiable on a neighbourhood of $c$, and that the limit $\lim_{x \to c} \frac{f'(x)}{g'(x)} = l$. In the proof we consider a neighbourhood of $c$, and we already know $f$ and $g$ are differentiable on the neighbourhood. We define $y$ to be a specific point such that $y > c$. Then $f$ and $g$ satisfy Cauchy's Mean Value Theorem on $[c,y]$.
We then know that $$\frac{f'(z)}{g'(z)} = \frac{f(y)-f(c)}{g(y)-g(c)} = \frac{f(y)}{g(y)}$$
So now we let $y \to c^+$ and it follows that $z \to c^+$.
By simply renaming the domain variable, we know that $$\lim_{z \to c^+} \frac{f'(z)}{g'(z)} = l$$
My book then says: It follows that $$\lim_{y \to c^+} \frac{f(y)}{g(y)} = l$$
This step is intuitively obvious to me; and my reasoning is that by the way we have defined $y$ and $c$, the expressions $\frac{f'(z)}{g'(z)}$ and $\frac{f(y)}{g(y)}$ are exactly equal for any corresponding values of $y$ and $c$, and thus the sequences defined by these two expressions are exactly equal. Thus the limits are obviously equal. Is this reasoning correct?
I also wanted to clarify how, if it is possible, I can take the limits of both the rhs and lhs of an equation when they are expressed using different variables? For example, if we have $f(x) = g(x)$, then I can simply say $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = m$. This is easy since there is only one domain variable. But in my particular case, how could I simultaneously take the limits of both sides of $\frac{f'(z)}{g'(z)} = \frac{f(y)}{g(y)}$? Is this even possible?
The only justification I can think of is the one I described above, where I deduce that the two sequences must be the same for the particular limits I have chosen, and thus must have equal limits. In other words, I don't know a general method for taking the limit of equivalent expressions of different variables?
 A: Those variables are different but not independent. We are taking limit of equation as a single variable $y\to c^+$. This forces $z\to c^+$ and the LHS to tend to $l$. By the nature of equality RHS also does the same.
You can use a typical $\epsilon, \delta$ to add more rigor if you wish. Thus let $\epsilon>0$ be given then we know that there is a $\delta>0$ such that $$0<|x-c|<\delta\implies |f'(x) /g'(x) - l|<\epsilon\tag{1}$$ Let $y$ be any number such that $0<y-c<\delta $. Then we have a $z$ with $c<z<y$ such that $$\frac{f'(z)} {g'(z)} =\frac {f(y)} {g(y)} $$ Now $c<z<y$ implies that $0<z-c<y-c<\delta$ and hence we automatically get from $(1)$ $$\left|\frac{f'(z)} {g'(z)} - l\right|<\epsilon $$ and therefore $$\left|\frac{f(y)} {g(y)} - l\right|<\epsilon $$ and the proof is complete.
Now consider the alternative scenario. Suppose the hypotheses are changed and you are given that $f(x) /g(x) \to l$ as $x\to c$. In this case we can't conclude that $f'(x) /g'(x) \to l$. Things change when we try to take limit of equation $$\frac{f'(z)} {g'(z)} =\frac{f(y) } {g(y)} $$ as $y\to c^{+} $. The variable $z\to c^{+} $ and RHS tends  to $l$ so does LHS. But we don't get the result $f'/g'\to l$. Why???
Because our entire argument uses the fact that for every $y$ there is some $z$ in $c<z<y$ for which the equation holds. Thus we can say safely that that there is some sequence of values $z_n$ with $z_n>c$ and $z_n\to c$ such that $f'(z_n) /g'(z_n) \to l$ but this does not work for all such sequences and hence $f'/g'$ may not necessarily tend to $l$.
The case of L'Hospital's Rule is different. Here we are given that for all sequences $z_n$ with $z_n\neq c, z_n\to c$ we have $f'(z_n) /g'(z_n) \to l$. And for all sequences $y_n$ with $y_n\neq c,y_n\to c$ we have some sequence $z_n$ with $z_n\neq c, z_n\to c$ such that $$\frac{f(y_n)} {g(y_n)} =\frac{f'(z_n)} {g'(z_n)} $$ It is known that RHS tends to $l$ and hence LHS tends to $l$ for all sequences $y_n$.
The key point is that there is an inherent difference between nature of $y$ and $z$ in your equation. We can find some $z$ for every $z$ which makes the equation true but we don't know if for every $z$ we can find a corresponding $y$. If that happens to be the case for some specific functions $f, g$ then L'Hospital's Rule will also work in reverse for those functions (if ratio of functions tends to a limit, then ratio of their derivatives tends to the same limit). 
