If a function #f(x)# has a vertical asymptote at #a#, then it has a asymptotic (infinite) discontinuity at #a#. In order to find asymptotic discontinuities, you would look for vertical asymptotes. Let us look at the following example.
#f(x)={x+1}/{(x+1)(x-2)}#
In order to have a vertical asymptote, the function has to display "blowing up" or "blowing down" behaviors. In the case of a rational function like #f(x)# here, it display such behaviors when the denominator becomes zero.
By setting the denominator equal to zero,
#(x+1)(x-2)=0 Rightarrow x=-1,2#
Now, we have a couple of candidates to consider. Let us make sure that there is a vertical asymptote there.
Is #x=-1# a vertical asymptote?
#lim_{x to -1}{(x+1)}/{(x+1)(x-2)}#
by cancelling out #(x+1)#'s,
#=lim_{x to -1}1/{x-2}=1/{1-2}=-1 ne pminfty#,
which means that #x=-1# is NOT a vertical asymptote.
Is #x=2# a vertical asymptote?
#lim_{x to 2^+}{x+1}/{(x+1)(x-2)}#
by cancelling out #(x+1)#'s,
#=lim_{x to 2^+}1/{x-2}=1/0^+=+infty#,
which means that #x=2# IS a vertical asymptote.
Hence, #f# has an asymptotic discontinuity at #x=2#.
I hope that this was helpful.
