How do you find the asymptotes for #y=1/(2-x)#?

1 Answer
Feb 16, 2016

Answer:

#y=0#
and
#x=2#

Explanation:

As #xrarr+-oo# the #2# in the denominator becomes less and less significant and
#lim_(xrarr+-oo) 1/(2-x) = 1/(+-oo) = 0#
Therefore #y=0# is an asymptote.

#y=1/(2-x# is defined for all values up to but not including #x=2#;
#yrarr +-oo# as #xrarr2#
Therefore #x=2# is an asymptote.

Here is what the graph looks like:
graph{1/(2-x) [-10, 10, -5, 5]}