Solutions

i)

Same limit definition, just different variable names.

ii)

It is correct in the sense that it implies the original definition. If we let $\alpha=\delta+1$, then

$$|f(x)-l|\leq \delta<\alpha.$$

iii)

Same as ii), except that if we let $\alpha=\delta/5$, then

$$|f(x)-l|<5\delta=\alpha.$$

iv)

Let $\alpha=\varepsilon/10$, and you have the original definition.