To apply L'Hôpital's Rule to the ratio which consists of the integrals Let us remember, the conditions to apply L'Hôpital's Rule:
Let suppose:
$f(x)$ and $g(x)$ are real and differentiable  for all $x\in (a,b)$ 
1-) $ \lim_{x\to c}{f(x)} = \lim_{x\to c}g(x) = 0$
2-)  If $g'(x)\neq 0$ $,\,\,\,\,$  as on some deleted neighborhood of $\,\,$ $c$.
3-)  $\,\,\,\,\,$ $\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L$, then
4-)  $\lim_{x\to c}{\frac{f(x)}{g(x)}}=L$, $\,\,\,\,\,$ thus we can write:
$$ \lim_{x\to c}{\frac{f(x)}{g(x)}}=\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L $$
Note 1: I have written all the above information just to remember to L'Hôpital's Rule. They are not related with my question. If I wrote something as incorect, you can correct it inside of the question. 
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ANYWAY, Let us suppose that our following example satisfies the conditions to apply L'Hôpital's Rule  which we stated above. If we start:
$\lim_{u\to \infty}\int_0^u F(t,a)dt=0$ $\,\,\,$  and$\,\,\,\,\,$ $\lim_{u\to \infty}\int_0^u G(t,a)dt=0$ 
$$\lim_{u \to \infty} \frac{ \int_0^u F(t,a) \, dt}{\int_0^u G(t,a) \, dt} $$ 
THE QUESTION. Before to apply L'Hôpital's Rule with respect to the parameter $´u´$ to the above-equation, do we need to also to prove the following one? 
There exists a number $M$ as $M> 0$.
The denominator $\,\,$ $\int_0^u G(t,a)dt$ $\,\,$ should not equal to zero for any$\,\,\,\,$ $u > M$.
Note 2: Please aware I am not asking anything about $\,\,$ $ \frac{d}{du} {\int_0^u G(t,a) \, dt}$ 
 A: Condition 2 needs two tweaks.  The first is that it doesn't make sense if $c$ is infinity.  Second, you don't need the denominator to be nonzero in the entire deleted neighborhood, but only in some neighborhood of $c.$ 
If you have a notion of "neighborhood of infinity," you can rewrite your condition as
2)  There is a deleted neighborhood about $c$ over which $g'(x)$ is nonzero.
Since you have $u\to \infty$, you need only that $G(u,a)$ (and it integral) are nonzero for all $u$ greater that some constant $M$.
There's another glitch in that your question seems to be about condition 2) which is about the derivative of the denominator, but when you get to your final question, you don't ask about the derivative(?)
A: The crucial issue here is the question of whether or not L'Hôpital's rule is applicable if all hypotheses are met except the possibility that $g'(x)$ can equal zero somewhere.

Where L'Hôpital's rule can fail -- for limits as $x \to +\infty$ -- is
  if for any $c > 0$ there exists $x_c > c$ such that $g'(x_c) = 0$.

A common misconception is that this is solely because the limit of $f'(x)/g'(x)$ cannot be defined if $g'(x)$ is zero infinitely often. This is not true because there can arise situations where $g'(x) = f'(x) h(x)$ with (1) $g'(x) = 0$ infinitely often, and (2) $h(x) \neq 0$ for all $x$, such that the limit of the ratio of derivatives can exist, i.e.,
$$\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{f'(x)}{f'(x)h(x)} = \lim_{x \to \infty} \frac{1}{h(x)}, $$
but $\lim_{x \to \infty} f(x)/g(x) $ can fail to exist.
As an example, take 
$$f(x) = \left(\int_0^x \cos^2 t \, dt \right)^{-1}, \quad g(x) = f(x) e^{\sin x}$$
Here we have $f(x), g(x) \to 0$ as $x \to \infty$, and 
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{1}{e^{\sin x}} \quad \text{DNE} $$
However,
$$\frac{f'(x)}{g'(x)} = \frac{-(f(x))^2 \cos^2x}{{-(f(x))^{2} \cos^2x \,e^{\sin x} + f(x) e^{\sin x} \cos x}} = \frac{\cos x}{e^{\sin x}(\cos x - (f(x))^{-1}) }$$
Since $e^{\sin x}$ is bounded between $e^{-1}$ and $e$ and $(f(x))^{-1} \to \infty$, we have 
$$\lim_{x \to \infty}\frac{f'(x)}{g'(x)} = 0$$
Hence,  L'Hôpital's rule fails here.  This is attributable to the fact that $g'(x)$ has the factor $\cos x$ which is zero for all $x= x_n = \frac{\pi}{2} + n\pi$.
