Answer
The statement is not sufficient to define any stable $L$ as the limit of $f(x)$ as $x$ approaches $c$.
Work Step by Step
"The number $L$ is the limit of $f(x)$ as $x$ approaches $c$ if, given any $\epsilon\gt0$, there exists a value of $x$ for which $|f(x)-L|\lt\epsilon$"
This is not enough to define limit unfortunately. Take the following example:
$$f(x)=3x-5$$
We know that $L=\lim_{x\to2}f(x)=3\times2-5=1$
Yet, how about we say that $L=\lim_{x\to2}f(x)=0$ and use the above statement to prove it? It is entirely possible to do so.
Take $L=0$, then $|f(x)-L|=|f(x)|$
- Given $\epsilon=1$, then we can choose $x=\frac{11}{6}$, then $|f(x)|=|3\times\frac{11}{6}-5|=|\frac{11}{2}-5|=|\frac{1}{2}|=\frac{1}{2}\lt\epsilon$
- Given $\epsilon=0.1$, we can choose $x=\frac{101}{60}$, then $|f(x)|=|3\times\frac{101}{60}-5|=|\frac{101}{20}-5|=|5.05-5|=|0.05|=0.05\lt\epsilon$
- Given $\epsilon=0.01$, we can choose $x=\frac{1001}{600}$, then $|f(x)|=|3\times\frac{1001}{600}-5|=|\frac{1001}{200}-5|=|5.005-5|=|0.005|=0.005\lt\epsilon$
And so on.
In other words, for any arbitrary value of $L$, given any $\epsilon\gt0$, we can always find a corresponding value of $x$ for which $|f(x)-L|\epsilon$, at least in the case above of $f(x)=3x-5$
Therefore, the statement above cannot be sufficient to define any stable $L$ as the limit of $f(x)$ as $x$ approaches $c$.
