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Why was the definition defines as $$ \forall \epsilon >0 \exists \delta \text{ such that if } 0<|x-a|< \delta \text{ then } |f(x)-L|< \epsilon$$ Rather than
$$\forall \epsilon >0 \exists \delta \text{ such that if } |f(x)-L|< \epsilon \text{ then } 0<|x-a|< \delta$$ The only reason I can think of is because $f(x)$ depends on $x$, but that isn't a satisfying answer since graphically it makes sense. That is as $f(x)$ gets closer to $L$, $x$ gets close to $a$. Secondly since it is for all $\epsilon$, it captures the idea of as "close as we want to $L$".

Side note What's confusing me here is that these two statements seem to capture the same idea which isn't the case.I'm hoping someone points out what's wrong with the second statement.

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    $\begingroup$ In the second definition, the a constant function never has any limits. You seem to be equating two statements $P \implies Q$ and $Q \implies P$, which is not valid. $\endgroup$
    – user296602
    Oct 19, 2017 at 20:30
  • $\begingroup$ I know I'm equating to statements that are not logically equivalent. However they both seem to have the same idea about limits which is why in confused. Can you please explain further what you mean by "the constant function never has a limit" $\endgroup$
    – Simple
    Oct 19, 2017 at 20:34
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    $\begingroup$ But they don't have the same idea at all. One says that similar outputs have similar inputs, and the other says that similar inputs have similar outputs. The fact that constant functions have the same output regardless of input shows why your second definition cannot capture what we mean by limits. $\endgroup$
    – user296602
    Oct 19, 2017 at 20:35
  • $\begingroup$ The second one can tell you there's a limit when there isn't one: For any gap, the second one only checks values for which $|f(x) - L| < \varepsilon$ is true in the first place, which is meaningless. The first definition is exactly what you need to say that every arbitrarily small open neighborhood around L is the image of some open neighborhood, which is the idea... $\endgroup$
    – dasaphro
    Oct 19, 2017 at 20:42
  • $\begingroup$ Dear @dasaphro, It is not the case that "every small open neighborhood around $L$ is the image of some open neighborhood." What is true is that every ball around $L$ contains the image of some ball around $a$. $\endgroup$
    – Keenan Kidwell
    Oct 19, 2017 at 20:56

1 Answer 1

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Consider the function $f$, defined by $$f(x) = \begin{cases} x+1, &x>0 \\ 0, &x=0 \\ x-1, &x<0\end{cases}$$

Clearly, it is discontinuous at $x=0$.

Now apply your second definition of continuity.

$\forall \varepsilon >0, \ \ \exists \delta>0$ such that $|f(x)-f(a)|<\varepsilon \implies |x-a|< \delta$.

I claim that for every $\varepsilon$, we can choose $\delta= \varepsilon$.

If $\varepsilon<1$, then $|f(x)-f(a)|<\varepsilon$ holds true only at $x=a$. So, $|x-a|<\delta$ is always true.

If $\varepsilon \geq 1$, then $|f(x)-f(a)|<\varepsilon$ is true when $x\in (-\varepsilon+1, \varepsilon-1) \subseteq (-\varepsilon,\varepsilon)$. So, $|x-a|<\delta$ is also true for this case.

Thus, your definition is satisfied for a discontinuous function.

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