An alternate definition of Limit Let's say we were to define the concept of limit as follows:
$\displaystyle{\lim_{x \to c}}f(x)=L$ means that for every $x$ in the domain of $f$, there exists an $x_0 \neq x$ in the domain of $f$ such that:
$$|x-c|>|x_0-c|$$and$$|f(x)-L|\ge|f(x_0)-L|$$
I have two questions:

*

*Does there exist a limit that can be proven using the (ε, δ)-definition but not by using the above definition?

*Does there exist a limit that can't be proven using the (ε, δ)-definition but can be proven using the above definition?


The issue is not what you can prove, but that these two definitions are different.

*

*Let us define the function:
$$
f(x)=\begin{cases}
1 & x=0\\
0 & x=1\\
x & x\ne 0 \text{ and } x\ne 1
\end{cases}
$$
Then, with the conventional definition, $\lim_{x\to 0} f(x)=0$. Take $x=1$. There is no point $x_0\ne x$ with $|f(x_0)|\le|f(x)|=0$. So the limit does not exist with your definition.


*Let $f(x)=\sin\dfrac{1}{x}$. For every $x\ne 0$, you can choose an integer $k$ large enough so that:
$$
x_0=\frac{1}{2\,k\,\pi}<|x|
$$
Then:
$$
|f(x_0)|=\left|f\left(\frac{1}{2\,k\,\pi}\right)\right|=|\sin 2\,k\,\pi|=0\le\left|\sin\dfrac{1}{x}\right|=|f(x)|
$$
This "proves" that $\lim_{x\to 0}f(x)=0$ with your definition. But the conventional limit does not exist.
ADDITION: As mentioned in other answers, your definition does not even define an unique limit. In the last example, choose $x_0=\frac{1}{2k\pi+m}$ with $\sin m=L$. Then, again $|f(x_0)-L|=0$ and $L$ would also be a limit for any $-1\le L\le 1$.
