Is the lim exists when #lim_(x->0) f(x)= +- oo?#

2 Answers
Oct 21, 2017

No, the limit does not exist.

Explanation:

Consider the example #f(x) =1/x#

#lim_(x->0^+) f(x) = +oo#

and

#lim_(x->0^-) f(x)= -oo#

Hence, one could conclude that #lim_(x->0) f(x) =+-oo#

However, since the limit tends to different results from above and from below the limit does not exist.

Dec 4, 2017

Many of us teach that so-called "infinite limits" are not limits that exist.

Explanation:

#lim_(xrarra) f(x) = L# is and only if for every positive #epsilon#, there is a positive #delta# such that, for all #x#, if #0 < abs(x-a) < delta#, then #f(x) - L) < epsilon#.

#L# cannot be a symbol that does not represent a number. The distance between #f(x)# and #L#, (#abs(f(x) -L)#) must be defined.

There are various ways that a limit can fail to exist.

One way is exemplified by #lim_(xrarr0) 1/x^2#.

In this case, the limit fails to exist because as #xrarr0#, #1/x^2# increases without bound.

We indicate this be writing #lim_(xrarr0)1/x^2 = oo#.

The definition is:

We write #lim_(xrarra) f(a) = oo# if and only if For any #M#, there is a delta such that is #0 < abs(x-a) < delta#, then #f(x) > M#.

Refusing to say that this limit exists allows us to simplify statements like:

If both #lim_(xrarra)f(x)# and #lim_(xrarra)g(x)# exist, then

#lim_(xrarra)f(x)g(x) = lim_(xrarra)f(x) lim_(xrarra)g(x)#.

This theorem fails if we include #lim_(xrarroo) f(x) = oo# as a limit that exists.