Why does $\epsilon = |L| + 2$ not work for this new definition of limit?

Question

Let's say that we were to change the standard definition of limit to "$\forall$ $\delta >0$ $\exists$ $\epsilon>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$." If we were to let $f(x) = \sin x$ and let $a$, $L$ and $\delta$ be arbitrary real numbers. Then why would $\epsilon = |L| + 2$ satisfying this new definition be a problem?

Answer 1

Your definition of "limit" actually says that $f$ is bounded in any open bounded interval. So $f = \sin$ does verify the condition.

About $\epsilon = |L| + 2$: you can't choose $\epsilon$. $\exists\epsilon$ means "there is some $\epsilon$", but not the $\epsilon$ that you want.