Confusion about $\epsilon - \delta$ language Suppose there is a function $f(x)$ and assume the limit exist as $x$ goes to $0$
$$\lim_{x\to0}f(x) = 0$$
then
$$\forall\varepsilon\gt0,\exists \delta\gt 0, \text{ such that } |x| \lt \delta ,|f(x)| \lt \varepsilon \tag{1}$$  
Can we also claim that, we can always find a $\delta_{1} \gt 0$ (an interval) such that
$$\forall\varepsilon\gt0,\exists \delta_{1}\gt0, \text{ such that } |x| \lt \delta_{1} ,|f(x)| \leq \varepsilon \tag{2}$$
Where the two $\varepsilon$ in $(1),(2)$ are the same. My question is: Given that $(1)$ holds, does $(2)$ also hold? Any help would be appreciated. Thanks in advance.
 A: It should be:  
$$\forall\varepsilon\gt0,\exists \delta\gt0,\quad\text{such that}\ \space |~x~| \lt \delta \Rightarrow ~|f(x)| \lt \varepsilon \tag{1}$$  
And yes, if $(1)$ holds then $(2)$ holds too. That's because in general $a \lt m$ implies $a \le m$.
If the two $\epsilon$ are the same, in $(2)$ just pick the same $\delta$. 
A: If (1) holds, then taking $\delta_1=\delta$ gives you (2).
If (2) holds, suppose we want to show (1) for a given $\epsilon$. By (2) there exists $\delta'_1$ such that $|x| < \delta'_1$ implies $|f(x)| \le \epsilon/2$. Then taking $\delta=\delta'_1$ gives the first definition since $|f(x)| \le \epsilon/2 < \epsilon$.

Comments: as Peter pointed out, there should be an implication "$\implies$" in (1) and (2). Also, I am assuming your question is mainly about the strict inequality vs the non-strict inequality. As user46944 points out, it is wrong to hope for a $\delta_1$ that works universally for any $\epsilon$. I answered your question under the assumption that $\delta_1$ depends on the chosen $\epsilon$.
