What is the limit of #x^n#?

1 Answer
Jul 13, 2015

#lim_(n->oo) x^n# behaves in seven different ways according to the value of #x#

Explanation:

If #x in (-oo, -1)# then as #n->oo#, #abs(x^n)->oo# monotonically, but alternates between positive and negative values. #x^n# does not have a limit as #n->oo#.

If #x = -1# then as #n->oo#, #x^n# alternates between #+-1#. So again, #x^n# does not have a limit as #n->oo#.

If #x in (-1, 0)# then #lim_(n->oo) x^n = 0#. The value of #x^n# alternates between positive and negative values but #abs(x^n) -> 0# is monotonically decreasing.

If #x = 0# then #lim_(n->oo) x^n = 0#. The value of #x^n# is constant #0# (at least for #n > 0#).

If #x in (0, 1)# then #lim_(n->oo) x^n = 0# The value of #x^n# is positive and #x^n -> 0# monotonically as #n->oo#.

If #x = 1# then #lim_(n->oo) x^n = 1#. The value of #x^n# is constant #1#.

If #x in (1, oo)# then as #n->oo#, then #x^n# is positive and #x^n->oo# monotonically. #x^n# does not have a limit as #n->oo#.