Implication in L'Hôpital's Rule When an unknown quotient $ \dfrac{0}{0} $ is indeterminate  it is hypothesized as an indeterminate maximum /minimum   point, is n't it?
For, when quotient of two functions
$ \dfrac{x}{y} = \dfrac{0}{0} = const.,$  is considered and we apply Quotient Rule for the first indeterminate fraction by differentiating w.r.t. $t$,
$$  \dfrac {y x^{\prime} - y^{\prime}  x}{y^2} = 0,  \quad  \dfrac {x}{y} =   \dfrac {x^{\prime} }{y^{\prime} }  \tag{1} $$
is what forms the L'Hôpital's Rule as a consequence. Right?
Now how far is the assumption of the unknown to be locally constant, and not be a variable,correct?
EDIT
Apologies. An error occurred while considering multi-variable calculus context, but here makes no sense.Not deleting it as there are already some answers. Sincere apology again for attentions gone improperly, prefer to close it in its present form.
 A: L'Hôpital is proven more or less like this: Say you have two continuously differentiable functions $f(x), g(x)$ such that $f(a) = g(a) = 0$, $g'(a) \neq 0$ and $\lim_{x \to a}\frac{f(x)}{g(x)}$ exists. (You can relax these constraints somewhat, specifically the "continuously differentiable" part, as well as allowing $g'(a) = 0$, if you want a more thorough proof, including the fact that repeated applications is valid. However, the basic idea is still the same.)
We have, for any $x$ where $\frac{f(x)}{g(x)}$ makes sense, that $\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)}$, since we're simply adding $0$ to both the numerator and the denominator.
But we're also allowed to divide the numerator and the denominator by any non-zero number we want. For instance, for the $x$-values for which $\frac{f(x)}{g(x)}$ make sense, we have that $x-a$ is non-zero. So
$$
\frac{f(x)}{g(x)} = \frac{f(x)- f(a)}{g(x) - g(a)} = \frac{\frac{f(x) - f(a)}{x-a}}{\frac{g(x) - g(a)}{x - a}}
$$
Lastly, since the leftmost and rightmost fractions are equal wherever they make sense, any of their limits must be the same as well. Specifically,
$$
\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{\frac{f(x) - f(a)}{x-a}}{\frac{g(x) - g(a)}{x - a}} = \frac{\lim_{x \to a}\frac{f(x) - f(a)}{x-a}}{\lim_{x \to a}\frac{g(x) - g(a)}{x - a}} = \frac{f'(a)}{g'(a)}
$$
A: One possible way to consider the problem : assuming that functions $f(x)$ and $g(x)$ have all required properties around $x=a$, use Taylor expansions $$f(x)=f(a)+(x-a) f'(a)+\frac{1}{2} (x-a)^2 f''(a)+\frac{1}{6} f'''(a)
   (x-a)^3+O\left((x-a)^4\right)$$ $$g(x)=g(a)+(x-a) g'(a)+\frac{1}{2} (x-a)^2 g''(a)+\frac{1}{6} g'''(a)
   (x-a)^3+O\left((x-a)^4\right)$$ and now, consider what happens to $$\frac{f(x)}{g(x)}=\frac{f(a)+(x-a) f'(a)+\frac{1}{2} (x-a)^2 f''(a)+\frac{1}{6} f'''(a)
   (x-a)^3+O\left((x-a)^4\right) }{ g(a)+(x-a) g'(a)+\frac{1}{2} (x-a)^2 g''(a)+\frac{1}{6} g'''(a)
   (x-a)^3+O\left((x-a)^4\right)}$$ when $f^{(n)}(a)$ and/or $g^{(m)}(a)$ tend to $0$.
