# Showing $\frac{d}{dx}\left(\frac{f(x)}{1 + cf(x)}\right) \rightarrow 0$ as $c \rightarrow \infty$

The problem I am working on is as follows:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuously differentiable, periodic of period 1, and nonnegative. Show that $$\frac{d}{dx}\left(\frac{f(x)}{1 + cf(x)}\right) \rightarrow 0$$ as $c \rightarrow \infty$ uniformly in $x$.

I can show it if $f$ is as above except instead of nonnegative, it is positive. The proof is as follows: Since $f$ is of period 1, consider $f$ on $[0, 1]$. Then $f$ attains a maximum $M$ and minimum $m > 0$ on $[0, 1]$. We also have $f'$ attaining a maximum at $\alpha$ on $[0, 1]$. Then $$|g_{n}(x)| = \left|\frac{d}{dx}\left(\frac{f(x)}{1 + nf(x)}\right)\right| \leq \frac{\alpha}{(1 + nm)^{2}} \rightarrow 0$$ as $n \rightarrow \infty$ (where the convergence is uniform in $x$). What should I do in the case when $f$ is nonnegative?

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It seems important that you use the fact that $f' = 0$ if $f=0$, which is where nonnegativity saves you (otherwise the statement is blatantly false). – Scaramouche Aug 19 '13 at 8:17
Thus, the natural thing to do would be to split the interval $[0,1]$ into two parts -- one where $f$ is away from 0 and the other where $f$ is close to 0. Where it is away from 0 you can bound as above, where it is close to 0 you use $C^1$ to get your bound. – Scaramouche Aug 19 '13 at 8:21