# How come the definition of a limit does not have $0\lt |f(x) - L| \lt \epsilon$ instead of $|f(x) -L|\lt \epsilon$?

We know that the definition of a limit is: $\forall \epsilon > 0 \space \exists\delta>0, 0<|x-c|<\delta \to |f(x) - L| < \epsilon$

From my understanding, this means that the closer $x$ gets to $c$, then the closer $f(x)$ gets to $L$. However, since $x$ is never actually at $c$, then isn't it impossible for $f(x) = L$? If so, then why don't write $0<|x-c|<\delta \to 0<|f(x) - L| < \epsilon$ since it $f(x)$ can never equal $L$?

It is possible for $f(x)$ to be $L$, even if $x\neq c$.

Maybe $f$ takes the value $L$ at another point than $c$. This happens for instance for all constant functions.

• Rephrased: If you would like constant functions to have limits you better use the standard definition. Sep 30, 2017 at 16:53

Is such an example what you are after: If $f(x) = 1$, then $f(x) = f(1)$ for all $x \neq 1$.

Let $f$ be the function such that

$$f(x)=\begin{cases} x\sin(1/x)&,x\ne0\\\\ C&,x=0 \end{cases}$$

where $C$ is an arbitrary number.

Then, we have $\lim_{x\to 0}f(x)=0$.

Note that $f(x)=0$ for an infinite number of values of $x$ on any deleted neighborhood of $x$ (i.e., just take $x=1/(n\pi)$ for any integer $n\ne 0$). And this is true despite the fact that $f(0)=C$ for any $C$.