Not necessarily. The use of L'Hopital's rule is essentially the application of a Taylor expansion for the numerator and denominator functions and considering the first non-zero term. Some smooth non-zero functions can have Taylor expansions where each and every coefficient is zero. For example $$f(x) = e^{-\frac{1}{x^{2}}}$$ at x=0 has Taylor coefficients of 0 to all orders, yet clearly if I consider g(x) = 2f(x) then the ratio $$\frac{f(x)}{g(x)}$$ at x=0 is simply 1/2, despite L'Hopital's rule being inapplicable yet the functions are both infinitely differentiable at x=0.
You should be careful, this is a bad counter example, you are subconsciously simplifying the fraction by g(x). g(x) does not exist in x=0.
The example you gave above is actually a classical example where the limit is undefined (not 1/2).