-1

The ϵ-δ definition of limits states:

Let $ƒ(x)$ be defined on an open interval about c, except possibly at $c$ itself. We say that the limit of $ƒ(x)$ as $x$ approaches c is the number $L,$ and write $\lim_{x \to c} f(x)=L$ if, for every number $\epsilon>0,$ there exists a corresponding number $\delta>0,$such that for all $x,$ $0<|x-c|<\delta$ implies $|f(x)-L|<\epsilon.$

My doubt is regarding the fact that why its written for every number $\epsilon>0,$ there exist a corresponding number $\delta>0$ and not the other way "for every number $\delta>0,$ there exist a corresponding number $\epsilon>0$" in this definition. Could someone please clarify my concepts?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
sab
  • 17
  • 1
    Consider $f(x)=\sin\frac1x$ at $c=0.$ Given any $\delta,$ we could pick $\epsilon=2$ and we could find the limit was any value in $(-1,1).$ – Thomas Andrews Apr 28 '24 at 12:36
  • Basically, in the correct definition, unless the function is constant, you need to pick smaller $\delta$ when you are given smaller $\epsilon,$ but in your definition, once we have an $\epsilon$ for one $\delta,$ that $\epsilon$ would work for all smaller values of $\delta.$ So it doesn't seem to match the intuition of "approaching $c.$" – Thomas Andrews Apr 28 '24 at 12:39
  • There are also some other similar questions linked from the one I linked. – Mark S. Apr 28 '24 at 13:20

2 Answers2

0

Suppose the function $f(x)$ defined as

$f(x) \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{otherwise} \end{cases}$

It's clear that it is not continuous at the point $x = 0 $. But if we define continuity as you suggested:

Let $ c = 0 $, and you give me $ \delta = a $, where $ a \in \mathbb{R} $, and I give you $ \epsilon = 2 $. It is true that:

$\forall x \in \mathbb{R}: |x-0| < a \rightarrow |f(x)- f(0)| < 2$

Then $f$ would be continuous at the point $0$.

Ulaş
  • 73
0

A way to think of the $\epsilon-\delta$ definition of the limit is "For any error tolerance $\epsilon$, we can specify $x$ to enough precision to get the value of $f(x)$ within that tolerance." Swapping these around says "For any level of precision of $x$, there is some error tolerance such that $f(x)$ is within that tolerance". I can always satisfy my error bounds if I'm allowed to choose the acceptable level of error after deciding the precision of $x$. So this doesn't say much.

eyeballfrog
  • 23,253