Why is it impossible to have #lim_(x->0) f(x)# and #lim_(f(x)->0)f(x)# simultaneously exist for any of these graphs?

#A)# #f(x) = 1/x^2#
#B)# #f(x) = -1/x^2#
#C)# #f(x) = 1/x#
#D)# #f(x) = -1/x#

2 Answers
Feb 12, 2017

All four of these graphs have the #x#-axis as a horizontal asymptote as #x-> pm infty# and the #y#-axis as a vertical asymptote as #x->0# from the right or left.

Feb 12, 2017

Well, by definition, a vertical asymptote is when at #x -> a#, #y -> pmoo# from either side and #x# never touches #a#. Similarly, a horizontal asymptote is when at #x -> pmoo#, #y -> b# from either side without ever reaching #b#.

For the function

#y = c/x#,

where #c# is a constant, if #x->0#, #y -> pmoo# from either side of #x = 0#, so you have a vertical asymptote. You can also find that as #x -> pmoo#, #y -> 0# but doesn't get there.

But if you have #x -> 0# and consequently #y -> pmoo#, you can't also have #x -> pmoo# so that #y -> 0#. It's not possible to approach both asymptotes at once because #x# cannot approach #0# and #pmoo# at the same time, and #y# cannot approach #pmoo# and #0# at the same time.

(Imagine trying to run to two different places at once; can't do it.)

Both kinds of asymptotes are on the graph, to be sure, but you can only approach one of those kinds of asymptotes at a time.