Show that $0 \le \max \{x,0\} \le \varepsilon $ if and only if $x \le 0$ 
Show that for any $x \in \Bbb R$   we have $0 \le \max \{x,0\} \le \varepsilon $ for all $\varepsilon \gt0$ 
  if and only if $x \le 0.$

How would I start an attempt at this question? 
I'm not sure how to start it, a hint would be appreciated!
 A: Hint: Assume $x>0$ by contradiciton and take $\epsilon = \frac{1}{2}x$.
A: The "if" part is trivial, since if $x \le 0$ then $\max\{x,0\} = 0$.
For the "only if" part, assume $ x > 0 $ so that $\max\{x,0\}=x$ and pick a value for $\epsilon$ with $0 \lt \epsilon \lt x$.
A: $$\max\{x, 0\} = \begin{cases}
x, & x > 0 \\
0, & x \leq 0\text{.}
\end{cases}$$
"$\Longrightarrow$" Suppose for all $\epsilon > 0$ that $0 \leq \max\{x, 0\} \leq \epsilon$. Suppose also, by contradiction, that $x > 0$. We wish to show that there is some value of $\epsilon$, say $\epsilon_0 > 0$, such that $\max\{x, 0\} \leq \epsilon_0$ isn't true, contradicting the assumption above. By the above, we have $\max\{x, 0\} = x$. 
Since $x > 0$, we may observe that for any $k \in (0, 1)$ that $kx > 0$ as well. Set $\epsilon_0 := kx$. Then
$$\max\{x, 0\} = x>kx$$
which is a contradiction.
"$\Longleftarrow$" Suppose $x \leq 0$. Then it follows that $\max\{x, 0\} = 0 \geq 0$ by the above, and for all $\epsilon > 0$, we have $\epsilon > 0 = \max\{x, 0\} \geq 0$. Since $\epsilon > \max\{x, 0\}$, $\epsilon \geq \max\{x, 0\}$ as well, hence $\epsilon \geq \max\{x, 0\} \geq 0$.
