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.