Is this a valid use of l'Hospital's Rule? Can it be used recursively? $$\lim_{x \to 0} \frac{1}{x} + \frac{1}{x}$$
$$\lim_{x \to 0} \frac{2x}{x^2}$$
Since this evaluates to an indeterminate form $\frac{0}{0}$ we use l'Hospital's Rule:
$$\lim_{x \to 0} \frac{2}{2x}$$
Since this also evaluates to an indeterminate form $\frac{2}{0}$ we use l'Hospital's Rule again:
$$\lim_{x \to 0} \frac{0}{2}=0$$
I know that I could have simply divided both numerator and denominator by $x$ to get the same result. This is just an example to ask the question: Can l'Hospital's Rule be used recursively?
EDIT: Sorry, I messed up my example. I cannot think of a good example right now but the question still stands. Can one use l'Hospital's Rule recursively?
 A: L'Hôpital's Rule
Assuming that the following conditions are true:


*

*$f(x)$ and $g(x)$ must be differentiable

*$\frac{d}{dx}g(x)\neq 0$

*$\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty}$


Then,
$$ \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \lim\limits_{x\to c} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}=L $$
Where $c$ and $L$ is any real number or $\pm\infty$.
So to answer your questions, yes, L'Hôpital's rule can be used repeatedly, provided that all of the above conditions are met. Since your example doesn't meet the aforementioned conditions, L'Hôpital's rule is not applicable.
Here is a case where L'Hôpital's rule is applicable multiple times,
$$ 
\lim_{x\to \infty}\dfrac{e^x}{x^2}=\frac{\infty}{\infty}
$$
Since the conditions are met, we can apply L'Hôpital's rule
$$
\lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}x^2}= \lim_{x\to \infty} \frac{e^x}{2x}= \frac{\infty}{\infty}
$$
Notice that the conditions are met again, so now
$$
\lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}2x}= \lim_{x\to \infty} \frac{e^x}{2}= \infty
$$
Therefore 
$$ 
\lim_{x\to \infty}\frac{e^x}{x^2}=\infty
$$
A: You're correct till the third step, although you can write $\dfrac1x+\dfrac1x = \dfrac2x$ instead of taking LCM.
However in the third step, $\lim\limits_{x\to0}\dfrac{2}{2x}$ is not an indeterminate form (not $\dfrac00$). Therefore, it's illegal inapplicable to use  L'Hospital's Rule there.
Answer to 'recursive' question: I don't know what you mean by 'recursively', but it can be used repetatively.
Consider $$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2} = \lim\limits_{x\to0}\dfrac{\sin x}{2x} = \lim\limits_{x\to0}\dfrac{\cos x}{2} = \dfrac12$$ by using  L'Hospital's Rule two times.
A: You can use the rule only if you are calculating $\lim \frac{f(x)}{g(x)}$ and $\lim f(x) = \lim g(x) = 0$ or $\lim f(x)=\lim g(x) = \infty$. The functions in $\frac2 {2x}$ do not satisfy this demand.
A: $\lim_{x\to 0} \frac{2}{2x}$ is not an indeterminate form.
A: I see what you did there.
You tried to claim that: $\lim_{x\to 0} \dfrac {2}{x} = \lim_{x\to 0} \left(\dfrac {2}{x}\times\dfrac{x}{x}\right)$ and as that was an indeterminant form you could apply l'Hospital's rule.
l'Hopital's only works if you cannot remove indeterminacy by cancellation.   It doesn't work if you introduce indeterminacy through multiplication.  
A: Yeah, but make sure to work the correct way around when writing it out. L'Hopital's rule says $f'(x)/g'(x)$ converges and $f(x)/g(x)$ is an indeterminate form, then $f(x)/g(x)$ has the same limit. So if you want to iterate it say twice and take advantage of $f''(x)/g''(x)$ being nice, you need to first apply it it $f'(x)/g'(x)$ by showing $f'(x)/g'(x)$ is an indeterminate form and $f''(x)/g''(x)$ converges to deduce $f'(x)/g'(x)$ converges, then you have enough information to apply the rule to to $f(x)/g(x)$.
