Is it ok to replace less with less than in (ε, δ)-definition of limit? Definition:

Let f(x) be a function defined on an interval that contains x=a,
except possibly at x=a . Then we say that,if for every number ε>0
there is some number δ>0 such that
$$\left| {f\left( x \right) - L} \right| < \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| < \delta$$

My question is whether I can change the def to the following
1):
$$\left| {f\left( x \right) - L} \right| <= \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| <= \delta$$
2):
$$\left| {f\left( x \right) - L} \right| < \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| <= \delta$$
3):
$$\left| {f\left( x \right) - L} \right| <= \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| < \delta$$
If not, could you please provide an counter-example? It seems all ok to me. But I am not certain.
 A: Given
$$\left| {f\left( x \right) - L} \right| <= \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| <= \delta$$
According to this def, since $\varepsilon$ is arbitrary small. Given $\varepsilon$ we can find a $\varepsilon_1$<$\varepsilon$ which also meet the condition. In this case:
$$\left| {f\left( x \right) - L} \right| < \varepsilon_1
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| <= \delta_1$$
When $0 < \left| {x - a}
\right| <= \delta_1$, the $\left| {f\left( x \right) - L} \right| < \varepsilon_1$ will always fullfill. So when pick a number $\delta_2$<$\delta_1$,$\left| {f\left( x \right) - L} \right| < \varepsilon_1$ still holds.
Thus we have:
$$\left| {f\left( x \right) - L} \right| < \varepsilon_1
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| < \delta_2$$
This could also be represented as
$$\left| {f\left( x \right) - L} \right| < \varepsilon
\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a}
\right| < \delta$$
