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?
Best Answer
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 $-1\le L\le 1$.