How prove $\lim_{x\to 0}f(x) = 0\iff\lim_{x\to 0}xf(x) = 0 $? Let $f:R \rightarrow R$ such that  $$| f(x+y)-f(x)-f(y) |\le |x-y|,$$ for all $x, y \in R.$ How can I prove that  $$\lim_{x\to 0}f(x) = 0\iff\lim_{x\to 0}xf(x) = 0? $$
 A: $(\implies)$ 
Straightforward. 
$(\impliedby)$

Edit: Following David C. Ullrich's comment, I turned the wlog assumption $f \geqslant 0$ into using $|f(x)|$ instead of $f(x)$ everywhere.

First note that substituting $x = y$ yields $|f(2x) - f(x) - f(x)| \leqslant 0$, so $f(2x) = 2f(x)$ for every $x \in \mathbb{R}$. 
Suppose that $\displaystyle \lim_{x \to 0} f(x) \neq 0$. We will show that also $\displaystyle \lim_{x \to 0} x \cdot f(x) \neq 0$. 
Exists such $\varepsilon > 0$ that there are arbitrarily small $x \neq 0$ with $|f(x)| \geqslant \varepsilon$. Let $\delta > 0$. There is such $x \neq 0$ that $|x| < \delta^2$ and $|f(x)| \geqslant \varepsilon$. There is a unique $n \in \mathbb{N}$ such that $\frac{\delta}{2} \leqslant |x| \cdot 2^n < \delta$ and then 
$$|f(x \cdot 2^n)| = 2^n \cdot |f(x)| \geqslant \frac{\delta}{2 |x|} \cdot \varepsilon > \frac{\delta}{2 \delta^2} \cdot \varepsilon = \frac{\varepsilon}{2 \delta}.$$
Thus $|x \cdot 2^n| < \delta$ and 
$$|x \cdot 2^n| \cdot |f(x \cdot 2^n)| \geqslant \frac{\delta}{2} \cdot \frac{\varepsilon}{2 \delta} = \frac{\varepsilon}{4}.$$
A: @David, I can't comment (not enough reputation), but $f(x)=1$ fails the first hypothesis at $x=y=1$.
A: From right to left
If $x=y$, we have
$$
|f(2x)-2f(x)|\le |x-x|\Longrightarrow f(2x)=2f(x)\Longrightarrow f(x)=ax,
$$
where $a$ is a constant, and statement is obvious.
