# Why is it $|f(x) - L| < \epsilon$ in the definition for a limit and not $0 < |f(x) - L| < \epsilon$?

In the epsilon-delta definition of a limit it says $0 < |x - a| < \delta$ must exist, i.e. the distance between $x$ and $a$ must be positive.

And then this leads to the implication that $|f(x) - L| < \epsilon$ holds where $\epsilon > 0$.

But why isn't it $0 < |f(x) - L| < \epsilon$? I thought limits were all about getting closer to a $y$ as you narrow in on an $x$. This definition implies that it is possible to bring $x$ near $a$ and yet somehow the limit $L$ is can possibly equal $f(x)$ as opposed to just getting closer and closer to it.

The definition intends to exclude the case $x=a$, so that a limit can be defined even if $f$ is undefined at $a$, as in say the definition of the derivative. In the case that the limit exists, it would be logically equivalent to replace $f$ (perhaps undefined at $a$) by some function $g$ which extends $f$ and is defined to have $g(a)=L$. But this is a bit backwards, so I think people avoid doing it.

On the other hand, it is completely permissible to have $f(x)=L$ even if $x$ is not equal to $a$. Think of the function $$f(x) = x\sin(1/x)$$ As $x \to 0$, no matter how small you choose $\delta$ there is an $x$ such that $0<|x|<\delta$ and $f(\delta)=0$. If you were to insist that $0<|f(x)|<\varepsilon$, then it would be false that $$\lim_{x \to 0} f(x) = 0$$ which is probably undesirable!

Said more formally: Say $f(x) \to L$ strongly if $$\forall \varepsilon >0 \quad \exists \delta>0 : 0 \le |x-a| <\delta \implies |f(x) - L| < \varepsilon$$ In other words, we change $0<|x-a|< \delta$ to just $|x-a|< \delta$. Then it is the case that if $f(x) \to L$ strongly, then $f(x) \to L$ in the usual sense. Moreover, if $f(x) \to L$ in the usual sense but $f$ doesn't tend to $L$ strongly, then the function $$g(x) = \cases{L \quad \quad x=a\\f(x) \quad \text{otherwise}}$$ tends to $L$ both strongly and in the usual sense. So the definitions are not too different.

On the other hand, say $f(x) \to L$ too strongly if: $$\forall \varepsilon >0 \quad \exists \delta >0 : 0<|x-a|<\delta \implies 0 < |f(x)-L| < \varepsilon$$ Now clearly, if $f(x) \to L$ too strongly then $f(x) \to L$ in the usual sense. But as we point out above, we exclude functions from the definition that we would prefer not to exclude.

• "On the other hand, it is completely permissible to have $f(x)=L$ even if $x$ is not equal to $a$." Does this only happen for periodic functions or functions where multiple $x$ values can yield the same $y$? Jan 27, 2018 at 21:15
• Would you accept $f(x) = x(1-x)$ for $x\ne 0$, and $f(x) = 1$ for $x =0$? Jan 27, 2018 at 21:18
• That function has $\lim_{x \to 0} f(x) = 0$, but equals $0$ only at $x=1$. Jan 27, 2018 at 21:19
• If $f$ is defined at $x=a$, and $f$ is continuous at $a$, and $f$ is injective (i.e. multiple values of $x$ cannot yield the same $y$), then $f$ cannot equal $L$ anywhere else. But there are many non-injective functions. Jan 27, 2018 at 21:21
• I understand that sometimes the limit at $x$ is not equal to $f(x)$, limits only talk about what happens around some $x$. But I am struggling to understand under what kind of scenarios would $f(x) = L$ exactly even when $|x-a| > 0$ Jan 27, 2018 at 21:21

With a limit you don't want to evaluate in $a$, so $0 < |x - a| < \delta$. Note that it does not say anything about $|f(x) - L|$ being non-zero.

Take for example $f(x)=c$ (or $f(x)=\frac{cx}{x}$), then $|f(x) - L|=0$ for all $x\neq 0$.

The function may take the limiting value at other points. The simplest example is if the function is constant. But what about $f(x) = x \sin (1/x)?$ When $x \rightarrow 0,$ the function takes the value $0$ at $1/n\pi,$ for all nonzero integers $n,$ and as you'll learn later in your course, the limit is 0.

Since $|f(x) - L|$ is non negative it is implied that $$0 \le |f(x) - L| < \epsilon$$

indeed $f(x)$ can also be equal to $L$.

Of course, $\lvert f(x)-L \rvert \ge 0$ always... And, equality is allowed...

Consider $(x_n)$ defined by $x_n=\begin{cases}\frac1n, n=2k\\0, n=2k+1\end{cases}$...