Construct a function which increases faster than its derivative. Is there a function $f$ which is differentiable in $(0, 1)$ and satisfies
\begin{equation*}
\begin{gathered}
\lim_{r \to 0+} |f(r)|=\infty, \\
\lim_{r \to 0+} \frac{|f(r)|}{|f’(r)|} = \infty.
\end{gathered}
\end{equation*}
 A: The question can be rephrased as: 

construct a smooth $f>0$ on $(0,1)$ such that $f(r)\to \infty$, $(\log f(r))'\to 0$ as $r\to 0$. 

Here, of course, the prime denotes derivative. Such a function cannot exist. If it existed, since $(\log f)'\to 0$ then the integral 
$$\tag{1}
\int_0^r (\log f(s))'\, ds$$
would be finite for all $r\in (0, 1)$. But direct computation shows that (1) equals 
$$
\log f(r) - \lim_{s\to 0} \log f(s) =-\infty.$$
A: We will deal with $r \to 1^-$. The same analysis applies to $r \to 0^+$ by replacing $f(r)$ with $f(1-r)$ and $f'(r)$ then with $-f'(1-r)$, which changes the sign of the limit in question.
Suppose that $f(r) \to \infty$, as $r \to 1^-$. Then there is some interval $(1-\epsilon, 1)$ on which $\log(f(r))$ is defined (because $f(r)$ is eventually positive) and $\log(f(r)) \to \infty$ as $r \to 1^-$ as well since $\log(s) \to \infty$ as $s \to \infty$. Then $$\frac{d}{dr}\log(f(r)) = \frac{f'(r)}{f(r)}$$ so
Pick a point $t_0 \in (1-\epsilon, 1)$. Then by the Mean Value Theorem, for each $r \in (t_0, 1)$, there is a point $c_r \in (t_0, r)$ such that $$\frac{d}{dr}\log(f)(c_r) = \frac{\log(f(r)) -\log(f(t_0))}{r - t_0}$$ and $r-t_0$ remains bounded away from zero as $r$ converges to $1$. Since $\log(f(r))$ converges to infinity as $r$ converges to the limit, we see that there is a sequence of points $x_n \in (t_0, 1)$ such that  $\frac{d}{dr}\log(f)(x_n) \to \infty$. Because $t_0 \in (1-\epsilon, 1)$ was arbitrary, this means that $$\limsup_{r\to1^-}\frac{f'(r)}{f(r)} = \limsup_{r\to1^-}\frac{d}{dr}\log(f)(r) = \infty.$$
This shows the question asked by the OP to be incorrect, because the limsup that we got applies to the reciprocal of the limit wondered about by the OP.
Moreover, we see that this result is the best that we can do, since we cannot guarantee convergence of this ratio to $\infty$, because $f'$ can get close to $-\infty$ while impacting $f(r)$ very little. For instance, $\sqrt{-x}$ on $(-1, 0)$ has derivative $\frac{-1}{2\sqrt{-x}} \to -\infty$ as $x \to 0$, but the change in the function over this finite interval is finite. Now, it seems that one can combine many of these examples and smoothly connect them to get a function such that $\limsup_{r\to1^-}\frac{f'(r)}{f(r)} = \infty$ and $\liminf_{r\to1^-}\frac{f'(r)}{f(r)} = -\infty$. In fact, this phenomenon of $f'$ being able to be close to $\infty$ in absolute value and $f$ changing very little is another reason that the limit wondered by the OP is false, because $f'$ can suddenly become very large, much larger than $f$, in small spurts and so the limit would not exist. 
Also, as a technicality, the denominator might equal zero at times if $f'(r) = 0$ often... but since you are wondering that the limit is infinite, then it really isn't a problem, I suppose.
A: The question was modified after I posted this answer. There is no such function: $f(x) >1$ and $\frac {f(x)} {f'(x)} >1$ for $0<x<r$ for some $r$. In particular $f'(x) >0$ for $0<x<r$ so $f$ is increasing in $(0,r)$. As $x$ decreases $f(x) $ decreases so it cannot have limit $\infty $ as $x \to 0$.
