Confusion regarding proof of L'Hopital rule. I was reading Spivak's Calculus. I have a query with regards to the proof for L'Hopital's rule for the 0/0 indeterminate form.
Theorem Statement :-
"Suppose that 
$
\lim_{x\to a} f(x) = 0 $ and 
$\lim_{x\to a} g(x) = 0 
$
and suppose that $ \lim_{x\to a} \frac {f'(x)}{g'(x)} $ exists .
Then $ \lim_{x\to a} \frac {f(x)}{g(x)} $ exists and
$$
\lim_{x\to a} \frac {f(x)}{g(x)} = \lim_{x\to a} \frac {f'(x)}{g'(x)}
$$
"
I have a confusion with regards to the last part of the proof. 
Using the Cauchy Mean Value Theorem it is shown that there exists a number $\alpha_x$ in $(a,x)$ such that $$
\ \frac {f(x)}{g(x)} = \ \frac {f'(\alpha_x)}{g'(\alpha_x)}
$$
Now $\alpha_x$ approaches $a$ as $x$ approaches $a$ because $\alpha_x$ is in $(a,x)$ , it follows that 
$$
\lim_{x\to a} \frac {f(x)}{g(x)} = \lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{y\to a} \frac {f'(y)}{g'(y)}
$$
My query is regarding the last two of the above equations. I understand that as $x$ approaches $a$ so does $\alpha_x$ , but then how are we treating $\alpha_x$ as a 'dummy variable' and replacing it with $y$. Is not $\alpha_x$ a dependent variable on x.
(I am also having some trouble in seeing how the step 
$$\lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)}$$ is justified )
Thanks in advance. I am new to these forums and looking forward to the discussion.
 A: The last two identities follow from the following fact:

Assume that the function $h$ has limit $\ell$ at $a$ and that the function $\alpha$ has limit $a$ at $x_0$, that is,
  $$
\lim_{t\to a}h(t)=\ell,\qquad\qquad\lim_{x\to x_0}\alpha(x)=a.
$$
  Then, the function $h\circ\alpha$ has limit $\ell$ at $x_0$, that is,
  $$
\lim_{x\to x_0}h(\alpha(x))=\ell.
$$

A simple proof is by the usual epsilon-delta approach.
A: These equalities hold because you assumed that the limit $\displaystyle\lim_{x\rightarrow a} \dfrac{f'(x)}{g'(x)}$ exists.  That means it doesn't really matter how you approach $a$ (i.e. either take $x\rightarrow a$ or $\alpha_x\rightarrow a$), you have to arrive at the same value for the limit.  This also explains why your last equality is okay, since you have some variable approaching $a$, it doesn't matter how you get there; essentially if the limit exists, then the variable respect to which you are taking the limit is automatically a dummy variable.
If you want to be more precise, you can use the $\varepsilon$-$\delta$ definition of the limit.  Say $L$ is your limit, which we assume exists.  Then for all $\varepsilon>0$ there exists some $\delta>0$ such that if $|a-x|<\delta$ then $\left|L-\dfrac{f'(x)}{g'(x)}\right|<\varepsilon$. But $|a-\alpha_x|<|a-x|<\delta$, so we know that for each $\varepsilon$, we can use the same $\delta$ to see that if $|x-a|<\delta$, then $\left|L-\dfrac{f'(\alpha_x)}{g'(\alpha_x)}\right|<\varepsilon$.  This immediately implies $\displaystyle \lim_{x\rightarrow a}\dfrac{f'(\alpha_x)}{g'(\alpha_x)} = L$.  
A: I see your point and it may be viewed as sloppy formulation.
We start on the left with an arbitrary sequence $x\to a$ and want to show that we obtain the same value for all such arbitrary sequences. Then we replace this arbitrary sequence with a specific sequence $\alpha_x\to a$ (depending on the given arbitray sequence). Since the premise is that the right hand side limit exists, i.e. that we obtain the same value for all sequences $y\to a$, we know that our specific sequence yields that very value.
