A curious problem with striking similarity to L'Hospital's Rule I got this problem in a book "A Problem Seminar" by Donald J. Newman which caught my attention. I am trying to find a solution and hence if possible please provide a hint only kind of solution.

Let $f, g$ be differentiable on $(0, 1]$ and $g'(x) > 0$ for all $x \in (0, 1]$ and further let $f'(x)/g'(x)$ tend to a limit (in the extended sense) as $x \to 0^{+}$. Prove that $f(x)/g(x)$ also tends to a limit (in the extended sense) as $x \to 0^{+}$.

By existence of limit in extended sense I mean that the limit is either a real number or is infinite but in any case the function does not oscillate.
Update: It appears that the requirement $g'(x) > 0$ is redundant. This is because $f'(x)/g'(x)$ is defined as $x \to 0^{+}$ and hence $g'(x)$ is non-zero and by intermediate value properly of derivatives $g'(x)$ maintains a constant sign as $x \to 0^{+}$.
Further update: The condition $g'(x) > 0$ is more for the non-vanishing of derivative $g'$ and this also ensures that $g$ is non-vanishing and hence both $g', g$ are of constant sign (via intermediate value property). We then have $(f/g)' = (g'/g)(f'/g' - f/g)$ and therefore if $f'/g' < f/g$ as $x \to 0^{+}$ or $f'/g' > f/g$ as $x \to 0^{+}$ then $f/g$ is monotone and hence it has a limit (in the extended sense).
The difficult case is when there are infinitely many values of $x$ as $x \to 0^{+}$ for which $f'/g' < f/g$ and there are infinitely many values of $x$ for which $f'/g' > f/g$. In this case I guess (but not sure) the $f/g$ remains near $f'/g'$ and the both tend to same limit. I wonder if this line of reasoning is correct. It appears that this reasoning is correct after all. See my answer.
Another reasoning goes like this. Since $g$ is monotone and of constant sign it follows that $g$ tends to a limit in the extended sense. If this limit is infinite then we know by a version of L'Hospital's Rule that $f/g$ tends to same limit as that of $f'/g'$. If $g$ tends to a finite limit (including $0$) then we need to show that in this case $f$ also tends to a limit in the extended and hence $f/g$ tends to a limit in extended sense.
 A: Finally I found a solution myself. I am assuming the following definition of limit of a function.
Let $f$ be defined in a certain deleted neighborhood of $a$. Then $f(x) \to L$ as $x \to a$ if for any arbitrary number $\epsilon > 0$ there is a number $\delta > 0$ such that $$|f(x) - L| < \epsilon$$ whenever $0 < |x - a| < \delta$.
Thus a pre-condition for existence of limit of $f(x)$ as $x \to a$ is that the function $f$ must be defined in a certain deleted neighborhood of $a$.
If we accept this definition then the constraint $g'(x) > 0$ is redundant in the question.
So I paraphrase the question:

Let $f, g$ be differentiable in $(0, 1]$ and let $f'(x)/g'(x)$ tend to a limit as $x \to 0^{+}$. Then prove that $f(x)/g(x)$ also tends to a limit as $x \to 0^{+}$.

Since $f'(x)/g'(x)$ tends to a limit as $x \to 0^{+}$ it follows that $g'(x)\neq 0$ when $x \to 0^{+}$ and by intermediate value property it means that $g'(x)$ is of constant sign as $x \to 0$. And therefore $g(x)$ must have constant sign as $x \to 0^{+}$. And the ratio $f(x)/g(x)$ is defined as $x \to 0^{+}$. We have $$\left(\frac{f(x)}{g(x)}\right)' = \frac{g'(x)}{g(x)}\left(\frac{f'(x)}{g'(x)} - \frac{f(x)}{g(x)}\right)$$ If the difference $$\frac{f'(x)}{g'(x)} - \frac{f(x)}{g(x)}$$ maintains a constant sign as $x \to 0^{+}$ then it means that $f(x)/g(x)$ is monotone and hence tends to a limit.
If this difference changes sign infinitely many times as $x \to 0^{+}$ it means that derivative of $f(x)/g(x)$ changes sign infinitely many times and thus we are led to maxima and minima of $f(x)/g(x)$ and at these points $f'(x)/g'(x) = f(x)/g(x)$. Since $f'(x)/g'(x)$ tends to a limit it follows that the values of $f(x)/g(x)$ at the critical points also tends to same limit and therefore all the intermediate value of $f(x)/g(x)$ tend to the same limit.
In all cases we can thus see that $f(x)/g(x)$ tends to a limit.
