Exercise 9 (Bonus) Let us recall that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous at $x_0$ if and only if for all $\varepsilon > 0$
there exists $\delta > 0$ such that for all $x$ such that $|x - x_0| < \delta$, we have $|f(x) - f(x_0)| < \varepsilon$.
Tell which one of the following statements are equivalent to the one above. If they are equivalent explain
why, if they are not find a counter-example.
1. For all $\varepsilon \ge 0$ there exists $\delta > 0$ such that for all $x$ such that $|x - x_0| < \delta$, we have $|f(x) - f(x_0)| < \varepsilon$.
2. For all $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x$ such that $|x - x_0| \le \delta$, we have $|f(x) - f(x_0)| < \varepsilon$.
3. For all $\varepsilon > 0$ there exists $\delta \ge 0$ such that for all $x$ such that $|x - x_0| \le \delta$, we have $|f(x) - f(x_0)| < \varepsilon$.
4. For all $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x$ such that $|x - x_0| < \delta$, we have $|f(x) - f(x_0)| \le \varepsilon$.
