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.