Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always? My question is summarised as follows:

Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$
or can we only deduce values of limits using differentiation in the forward direction, ie we can't deduce that the value of $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ is the same as $\lim_{x\to a}\frac{f(x)}{g(x)} $?

Logically I cannot see why this shouldn't be true (ie I can't see why we can't also integrate the numerator and denominator). HOWEVER, the following result seems to throw some doubt on this:
Consider the identity
$$f(x)^2\frac{d}{dx}\frac{g(x)}{f(x)}+g(x)^2\frac{d}{dx}\frac{f(x)}{g(x)}\equiv0$$
$$\implies\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv-\frac{f(x)^2}{g(x)^2}$$
On applying L'Hopital's rule we obtain
$$\lim_{x\to a}\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}\implies\lim_{x\to a}\frac{\frac{f(x)}{g(x)}}{\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
$$\implies\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
which is obviously false.
Any help explaining this apparent proof would be appreciated.
 A: The answer to your first question is No.
Take $$f(x)=x^2\sin(\frac 1x)$$
and
$$g(x)=x$$
then
$$\lim_{x\to 0}\frac{f(x)}{g(x)}=0$$
But
$$\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
Does not exist.
A: For clarity, let's let $F(x)=f(x)/g(x)$ and $G(x)=g(x)/f(x)$. The identity you call attention to can be written as
$${F'(x)\over G'(x)}=-{f(x)^2\over g(x)^2}$$
and we also, of course, have
$${f(x)^2\over g(x)^2}={f(x)/g(x)\over g(x)/f(x)}={F(x)\over G(x)}$$
Now if L'Hopital's rule applied to the limit for $F/G$, then we would have
$$\lim{f(x)^2\over g(x)^2}=\lim{F(x)\over G(x)}\color{red}{=}\lim{F'(x)\over G'(x)}=-\lim{f(x)^2\over g(x)^2}$$
(where the red equal sign is the L'Hopital step).  But in order to apply L'Hopital to the ratio $F/G$, we must have either $\lim F(x)=\lim G(x)=0$ or $\lim F(x)=\lim G(x)=\pm\infty$, and this cannot happen, since $F(x)$ and $G(x)$ are reciprocals (i.e., $G(x)=1/F(x)$). In short, there is no paradox, because L'Hopital does not apply to the ratio $F/G$.
A: The standard theorem says that both limits are equal if

*

*$f$ and $g$ are differentiable on an interval containing $a$, with the possible exception of $a$ itself

*$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} |g(x)| = \infty$

*$\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists.

You can use it to find the limit of $f(x)/g(x)$ from the limit of $f'(x)/g'(x)$ or to find the limit of $f'(x)/g'(x)$ from the limit of $f(x)/g(x)$, but you do need to know that the limit of $f'(x)/g'(x)$ exists.
EDIT: Your "apparent proof" is bogus because it violates assumption (2).
A: 
Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$


I can't see why we can't also integrate the numerator and denominator

As other answers have already noted, these limits are not always equal (even when they exist).
Let’s try to find the simplest possible explanation here. We start by noting that differentiation produces only one answer, but integration (in the sense of antidifferentiation) produces infinitely many answers, expressed using a “constant of integration”. Changing this constant does not affect the first limit in the question, but we expect that it does affect the second.
Let $f_1$ and $f_2$ be functions where $f_2(x) = f_1(x) + c$ for some constant $c \neq 0$. Then, assuming that the relevant limits exist and are finite,
\begin{align*}
\lim_{x \to a} \frac{f_2(x)}{g(x)} &= \lim_{x \to a} \frac{f_1(x) + c}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + \lim_{x \to a} \frac{c}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + c \cdot \lim_{x \to a} \frac{1}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + c \cdot \frac{1}{\lim_{x \to a} g(x)} \\
&\neq \lim_{x \to a} \frac{f_1(x)}{g(x)}
\end{align*}
So we were right: the second limit changes (in at least some cases). Therefore, the limits are not always equal.
