L'Hopitals Rule and division by zero? My question concerns division by zero. Let's say there are two functions, $g(x)$ and $f(x)$ that approach $0$, as $x\to t$, also assume their derivative w.r.t. $x$ is finite as $x\to t$. Using L'Hopital's Rule: $\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$ $x\to t$, Can you say $\frac{f'(x)}{f(x)} = \frac{g'(x)}{g(x)}$? Or $\frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} = 0$? To be more specific is $\frac{1}{0} = \frac{1}{0}$? Or is it undefined?
 A: When you talk of $f(x)/g(x)$ then you assume that $g(x) \neq 0$ as $x \to t$. At the same time you assume that $f(x) \to 0, g(x) \to 0$ as $x \to t$. This is the typical use-case where L'Hopital's Rule can be applied provided $f'(x)/g'(x)$ tends to a limit as $x \to t$.
Now note that if $f'(x)/g'(x) \to L$ as $x \to t$ then it is also obvious that $g'(x) \neq 0$ as $x \to t$. The L'Hopital's Rule says that under these conditions $f(x)/g(x) \to L$ as $x \to t$. This means that $$\lim_{x to t}\frac{f(x)}{g(x)} = \lim_{x \to t}\frac{f'(x)}{g'(x)}\text{  or what is the same as  }\lim_{x \to t}\left(\frac{f(x)}{g(x)} - \frac{f'(x)}{g'(x)}\right) = 0\tag{1}$$ but this does not mean that $$\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$$ Note further that if two functions have same limit as $x \to t$ then it does not necessarily mean that they are equal as $x \to t$.
The symbols like $1/0$ don't mean anything and it is better not to worry about them.
A: You cannot say any of those things. Division by $0$ is ALWAYS undefined (in the real numbers before anyone says anything).
A: Arguably, $\frac10$ can take on any or no value. Therefore it is incorrect, though this may seem strange, to say that $\frac10=\frac10$ in any (real) case. Of course, any value approaching zero but not zero can be compared.
Edit: In any other real case, $\frac1x=\frac1x$, because values only have a single reciprocal.
