# Connection between $\epsilon-\delta$ definition of limit and Weierstrass definition of continuous functions

I have just read the definition of a continous function according to Weierstrass, and for me it seems the same definition of the $\epsilon-\delta$ definition for limits.

I also know that there exists a definition of continuity of a function at some point in its domain by using the definition of limits, so it seems that Weierstrass simply tried to cheat everyone saying that his definition is different from the one that says that:

A function $f: I \rightarrow \mathbb{R}$ is continous at a point $c \in I$ if $$\lim_{x\to c}f(x) = f(c)$$

In general it seems that Weierstrass definition tries to be more independent of limits, but under the hood it seems to be the same or at least to go in the same way..

So is this well known or am I wrong?

• Is it Weierstrass whose character you are assassinating? – muaddib Jun 10 '15 at 19:43
• @muaddib ahah no, but I thought he simply introduced more complexity where it was not necessary... – nbro Jun 10 '15 at 19:45
• Gotcha, I'll be interesting in seeing the answers. – muaddib Jun 10 '15 at 19:49

Unless I'm misunderstanding which definition of Weierstrass you're referring to, yes: They are equivalent. The advantage of the $\epsilon-\delta$ definition is that it is more explicit; it simply unravels the statement $\lim_{x \to c}f(x) = f(c)$ in terms of the definition of the limit.
Suppose $I$ is an interval and $f: I \to \mathbb{R}$ is a function. Then $f$ is continuous at $c \in I$ if and only if $\lim_{x \to c}f(x) = f(c)$. By the definition of the limit, $\lim_{x\to c}f(x) = f(c)$ if and only if for all $\epsilon > 0$, there exists $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$.