What are all the removable discontinuities/holes and vertical asymptotres for #(2x^2)/(x^2-1)#?

1 Answer
Oct 26, 2015

Since #x^2-1=(x+1)(x-1)# we will have discontinuities
at #x=+1andx=-1# (or the denominator will be #=0#

Explanation:

These discontinuities are non-removable, as it makes a difference whether you approach these values from below of from above (try this and see the graph). This means that #x=+1andx=-1# are vertical asymptotes.

Also, when #x# grows bigger (positive or negative) the function starts to look more like
#(2cancel(x^2))/(cancel(x^2))=2# so #y=2# is a horizontal asymptote.
Between the values of #x=+1andx=-1# the function will be negative with a maximum at #(0,0)#, so in the values of the function there is a gap between #0and+2#
graph{2x^2/(x^2-1) [-16.02, 16.01, -8.01, 8.01]}