# Does the value of a function at a point have to exist in order for the limit to exist at that point?

Jul 31, 2018

No, it does not.

#### Explanation:

Let $f \left(x\right)$ be a real function of real variable defined in the domain $I \subset \mathbb{R}$.

Let now $\overline{x} \in \mathbb{R}$ be a generic real number.

If $\overline{x} \in I$ then the function has a value in that point: $f \left(\overline{x}\right)$.

On the other hand, the limit:

${\lim}_{x \to \overline{x}} f \left(x\right)$

is defined if $x$ is a point of accumulation for $I$, which means that in every interval $\left(a , b\right)$ such that $\overline{x} \in \left(a , b\right)$ we can find a point $\xi \in I \cap \left(a , b\right)$. This is possible also if $\overline{x} \notin I$

For example consider the function:

$f \left(x\right) = \ln \frac{1 + x}{x}$

Here $I = \left(- 1 , 0\right) \cup \left(0 , + \infty\right)$, so the function is not defined for $x = 0$. Hower $x = 0$ is a point of accumulation for $I$ because in any interval $\left(- \delta , \delta\right)$ we can find points where $f \left(x\right)$ is defined. (actually all of them except $x = 0$).

In fact:

${\lim}_{x \to 0} \ln \frac{1 + x}{x} = 1$

A more extreme example is to consider a function $\phi \left(q\right)$ defined only for rational numbers, $q \in \mathbb{Q}$.

We could still evaluate ${\lim}_{q \to \sqrt{2}} \phi \left(q\right)$, because though $\sqrt{2}$ is irrational, it can be approximated with a rational number as closely as we want.

By the way the converse is also true. The limit:

${\lim}_{x \to \overline{x}} f \left(x\right)$

cannot be defined even if $\overline{x} \in I$, if $\overline{x}$ is not a point of accumulation for $I$.