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)# 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#.