The graph of $f(x)$ lies below this line for small values of $x$

Question

Suppose that there exists a function $f:\mathbb R_{+} \to \mathbb R_{+}$ with only the following properties :

1. $f(x)>0$ for every $x>0$
2. $\lim_{x \to 0} f(x) =0$

My question: Are $(1),(2)$ sufficient in order to deduce that $f(x) \leq 2x\;$ if $x$ is sufficiently small?

My approach: I tried with the standard $\epsilon$-$\delta$ definition, i.e $\begin{align} \forall \epsilon>0 \quad \exists \delta>0 \;\text{ such that } \; 0<x<\delta \implies f(x)<\epsilon\;. \end{align}$ Then, choosing $\delta=\frac{\epsilon}{2}$, I obtain that $f(x) <2\delta\;$ which then contradicts what I need to show. Does this mean that my question has a negative answer or am I missing something here?

Any help is much appreciated! Thanks in advance.

Answer 1

There is the counter example $f(x)=3x$, which satisfies both properties ($3x>0$ for every $x>0$ and $\lim_{x \to 0} 3x =0$, because $f(0)=0$). It is also always bigger than $2x$.

The answer is that those properties aren‘t sufficient to deduce that $f(x) < 2x$ for sufficiently small values of x.