prove or disprove: $\lim_{x\to \infty} \frac{f(x)}{g(x)}=\lim \frac{f'(x)}{g'(x)}=0 \implies \lim \frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}}\ne 0$ my attempt $$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}}\text{ yields }\frac{0}{0}$$ then use l'hopital on this 
$$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}} =\lim_{x\to \infty} \frac{\frac{f'}{g}-\frac{fg'}{g^2}}{\frac{f''}{g'}-\frac{f'g''}{g'^2}} $$
$$\lim_{x\to \infty}\frac{\frac{f(x)}{g(x)}}{\frac{f'(x)}{g'(x)}} =\lim_{x\to \infty} \frac{\frac{f'}{f}\frac{f}{g}-\frac{f}{g}\frac{g'}{g}}{\frac{f''}{f'}\frac{f'}{g'}-\frac{f'}{g'}\frac{g''}{g'}} $$
$$\lim_{x\to \infty}\frac{\frac{f}{g}}{\frac{f'}{g'}} =\lim_{x\to \infty} \frac{\frac{f}{g} \left( \frac{f'}{g'}-\frac{f}{g}\right)}{\frac{f'}{g'} \left( \frac{f''}{f'}-\frac{g''}{g'}\right)} $$ assuming such limits exists, and equal to $L$,  leads to 
$$\lim_{x\to \infty}\frac{\frac{f}{g}}{\frac{f'}{g'}} =\lim_{x\to \infty} \frac{\frac{f}{g}  }{\frac{f'}{g'}  }\cdot \lim \frac{  \left( \frac{f'}{g'}-\frac{f}{g}\right)}{ \left( \frac{f''}{f'}-\frac{g''}{g'}\right)} $$
$$L =L\cdot \lim_{x\to \infty} \frac{  \left( \frac{f'}{g'}-\frac{f}{g}\right)}{ \left( \frac{f''}{f'}-\frac{g''}{g'}\right)}  $$
I am stuck here.. 
I would really like to prove that $L$ is not zero and not $\infty$ the proof is  obvious for polynomials functions f, and g, and I can not find any counter examples.. any help would be much appreciated..  
we can assume the initial condition comes from appropriately using l'hopital
 A: Let $f(x)=e^{-x^2}$ and $g(x)=e^{-x}$.
A: Just saw Barry Cipra had given the same answer...not sure how I could have missed that.
If $f=e^{-x^2}$ and $g=e^{-x}$, you get
$$
\lim_{x\rightarrow\infty}\frac{f}{g}
=\lim_{x\rightarrow\infty}\frac{f'}{g'}
=\lim_{x\rightarrow\infty}\frac{f/f'}{g/g'}
=0
$$
which is a counter-example to your hypothesis.
For the alternative hypothesis ($L=\infty$), you can set $f=e^x$, $g=e^{x^2}$ to get
$$
\lim_{x\rightarrow\infty}\frac{f}{g}
=\lim_{x\rightarrow\infty}\frac{f'}{g'}
=0,
\quad
\lim_{x\rightarrow\infty}\frac{f/f'}{g/g'}
=\infty.
$$
A: This seems like it would require some type of lemma which described the behavior of the products and quotients of $\ f:f'::f':f''$.  You can factor $\ f'/g'$  out of the expression you stopped at, but are then stuck with a difference quotient with unlike terms.  Without a rigorous lemma which would allow you replace limits of the divisor and dividend with known quantities, it seems unlikely you would be able to simplify further.  I'll post more if I come up with something further.  Interesting post and concept, however.  
One differing thought is that it might be necessary to first prove that indeterminate form is never zero, but I think that is a definition and not provable because indeterminate form has logical elements that correspond to defined numbers and ones that are always undefined (i.e., $\ n/0 $).
Regarding some of the answers ... Let statements can disprove the hypothesis, but the question remains as to "why" they do so.  In the above let statements, the reason the hypothesis can be disproved is because you began with the supposition that indeterminate form can be determined not to be zero.
