If for every positive number $\varepsilon$ there exists a positive number $\delta$, why not the other way back?

Question

Definition.

Let $f(x)$ be a function of $x$. If for every positive number $\varepsilon$, however small it may be, there exists a number $\delta$ such that whenever 0 $<$ |x - a| $<$ $\delta$ we have |f(x) - l| $<$ $\varepsilon$, then we say $f(x)$ tends to the limit $l$ as 'x tends to a' and write

$\lim_{x \to a}$ $f(x)$ = $l$

Proposed definition.

Let $f(x)$ be a function of $x$. If for every positive number $\delta$, however small it may be, there exists a number $\varepsilon$ such that whenever 0 $<$ |x - a| $<$ $\delta$ we have |f(x) - l| $<$ $\varepsilon$, then we say $f(x)$ tends to the limit $l$ as 'x tends to a' and write

$\lim_{x \to a}$ $f(x)$ = $l$

Question: How would the proposed definition affect the way we evaluate a limit?

Answer 1

Consider the function $$ f(x) = \begin{cases} 1,&\text{if } x>0, \\ 0,&\text{if } x\le0.\end{cases} $$ This function is clearly not continuous at $x=0$; from the $\varepsilon$-$\delta$ definition, we can see this by choosing $\varepsilon=\frac12$, and noting that there does not exist $\delta>0$ such that $|f(x)|=|f(x)-f(0)|<\frac12$ for all $|x|<\delta$.

However, this function does satisfy a hypothetical "$\delta$-$\varepsilon$ definition": for any $\delta>0$, we can choose $\varepsilon=2$, and it is definitely true that $|f(x)|<\varepsilon$ whenever $|x|<\delta$. (Indeed, any bounded function would satisfy this "$\delta$-$\varepsilon$ definition" for example, and there are plenty of bounded functions that are not continuous.)

(I hope this answers the question you mean to ask. The statement "$\varepsilon$ implies $\delta$" doesn't have a precise meaning when $\varepsilon$ and $\delta$ are numbers, and I'd encourage you while learning the subject to make sure that you use the mathematical notation in precise ways—doing otherwise allows misconceptions to become more entrenched.)

Answer 2

You proposed definition

Let $f(x)$ be a function of $x$. If for every positive number $\delta$, however small it may be, there exists a number $\varepsilon$ such that whenever 0 $<$ |x - a| $<$ $\delta$ we have |f(x) - l| $<$ $\varepsilon$, then we say $f(x)$ tends to the limit $l$ as 'x tends to a' and write $\lim_{x \to a}$ $f(x)$ = $l$

is a valid definition, but not for a limit. In fact, the bolded phrase is misleading because if it is true for a single $\,\delta>0\,$ then it implies it is true for all smaller $\,\delta>0.\,$ Thus, it would be better phrased with small replaced with large. What the definition defines is that the function $\,f(x)\,$ is bounded in any deleted neighborhood of the real $\,a\,$ and furthermore, the value of the real $\,l\,$ does not matter since we can always choose $\,\varepsilon\,$ big enough so that the interval $\,(l-\varepsilon,l+\varepsilon)\,$ contains the interval $\,[\liminf(f(x)),\limsup(f(x))]\,$ of the range of $\,f(x)\,$ where $\,x\,$ is restricted to $\, 0 < |x - a| < \delta.\,$ In fact, the real $\,a\,$ does not matter for the same reason.

Also, there is now no need to restrict to a deleted neighborhood. The restriction should be to $\, |x - a| < \delta\,$ instead because the value $\,f(a)\,$ does not cause any problems since it only requires that the bounds include it. Its value does not change the boundedness of the function. Note that in the actual definition of continuity, you forgot to explictly state that both $\,a\,$ and $\,l\,$ are given reals. An improved version of your definition for a locally bounded function with the role of $\,a\,$ and $\,l\,$ made explicit is:

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Let $f(x)$ be a function of $x$. If for every real $\,a\,$ and positive real $\delta$, however large it may be, there exists reals $\varepsilon$ and $\,l\,$ such that whenever $|x - a| < \delta$ we have $|f(x) - l| < \varepsilon,\,$ then we say that $\,f(x)\,$ is a locally bounded function.

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