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

1 Answer
Jul 31, 2018

No, it does not.

Explanation:

Let f(x)f(x) be a real function of real variable defined in the domain I sub RR.

Let now barx in RR be a generic real number.

If bar x in I then the function has a value in that point: f(bar x).

On the other hand, the limit:

lim_(x->barx) f(x)

is defined if x is a point of accumulation for I, which means that in every interval (a,b) such that barx in (a,b) we can find a point xi in I nn (a,b). This is possible also if barx notin I

For example consider the function:

f(x) = ln(1+x)/x

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

In fact:

lim_(x->0) ln(1+x)/x = 1

A more extreme example is to consider a function phi(q) defined only for rational numbers, q in QQ.

We could still evaluate lim_(q->sqrt2) phi(q), because though sqrt2 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->barx) f(x)

cannot be defined even if bar x in I, if barx is not a point of accumulation for I.