An alternate definition of Limit

Question

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:

1. Does there exist a limit that can be proven using the (ε, δ)-definition but not by using the above definition?
2. Does there exist a limit that can't be proven using the (ε, δ)-definition but can be proven using the above definition?

Answer 1

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

1. 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.

2. 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 $-1\le L\le 1$.

Answer 2

$L$ can even not be in the range of $f$.

Take for instance $f(x)=1_{\mathbb Q}-1_{\mathbb R\setminus\mathbb Q}$ and $L=0$.

We have $f(x)=\pm 1$ so $f$ cannot converge in traditional sense, but since $|f(x)-L|=\text{cst}$ and you have set for a loose inequality, your condition is always fulfilled.

This is a tortured function but even take $f(x)=\ell$ then by your definition $\lim\limits_{x\to c}f(x)=2\ell$ (or any $u\in\mathbb R$ in fact). That fact that constant functions converge to any limit is kind of "annoying".

Answer 3

The first requirement can't be satisfied unless you specify "for every $x\neq c$ in the domain of $f$".

Examples of limits that exist in the $\epsilon,\delta$ definition but not this definition:

1. Consider $f(x)=x^2$. In the new definition, for any $L<0$, we have $\lim_{x\rightarrow 0}f(x)=L$, since for any $x\neq 0$, we can choose $x_0$ such that $\vert x_0\vert<\vert x\vert$, and then $\vert x^2-L\vert > \vert x_0^2-L\vert$. This also shows that limits are not uniquely defined with this definition.

2. For the function $f(x)=2x^2-4x^4$, there are two global maxima at $x=\pm \frac{1}{2}$. Then $\lim_{x\rightarrow \frac{1}{2}}$ does not exist in this definition. For all $L<f(\frac{1}{2})=\frac{1}{4}$, we can take $x$ to be the preimage of $L$ under $f$ that is closest to $c=\frac{1}{2}$. Then $\vert f(x)-L\vert=0$, but $f(x_0)\neq L$ for any $x_0$ closer to $c$ than $x$, so $L$ cannot be the limit

   For $L\geq \frac{1}{4}$, take $x=-\frac{1}{2}$. Then since $f(x_0)<\frac{1}{4}$ for all $x_0\neq \pm\frac{1}{2}$, this means $\vert f(x)-L\vert <\vert f(x_0)-L\vert$ for all $x_0\neq x,c$. Hence, $L$ cannot be the limit.

More generally, limits won't be defined for any points where the function is not injective, by the same logic as (2).