Question 625894
<pre>

f(x)={{{(x-2)/(x^2-3x+2)}}}

Here is the graph:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10,(x-2)/(x^2-3x+2)),circle(2,1,.2),green(line(1,15,1,-15)) )}}}



{{{matrix(2,2,

lim, f(x),
"x->1","" )}}} does not exist because

{{{matrix(2,2,

lim, f(x),
"x->1","" )}}} = {{{matrix(2,1,

lim, 
"x->1" )}}}{{{(matrix(1,1,(1-2)/(1^2-3*1+2)))}}} = {{{cross(-1/0)}}}

Therefore lim, f(x),
"x->1","" )}}} does not exist because the denominator becomes 0 but the
numerator does not become 0, but becomes -1. Therefore, there is a
non-removable infinite discontinuity at x=1. That's where ther is a
vertical asymptote. 

{{{matrix(2,2,

lim, f(x),
"x->2","" )}}} = {{{matrix(2,1,

lim, 
"x->2" )}}}{{{(matrix(1,1,(2-2)/(2^2-3*2+2)))}}} = {{{cross(0/0)}}} 


There is no point at (2,1). However since both numerator and
denominator approach 0, the limit might exist and there may be a
removable discontinutity there indicated by the small circle on
the graph.

{{{matrix(2,2,

lim, f(x),
"x->2","" )}}} = {{{matrix(2,1,

lim, 
"x->2" )}}}{{{(matrix(1,1,(x-2)/(x^2-3x+2)))}}} = {{{matrix(2,1,

lim, 
"x->2" )}}}{{{(matrix(1,1,(x-2)/((x-2)(x-1))))}}} = {{{matrix(2,1,

lim, 
"x->2" )}}}{{{(matrix(1,1,(cross(x-2))/((cross(x-2))(x-1))))}}} = {{{matrix(2,1,

lim, 
"x->2" )}}}{{{(matrix(1,1,1/(x-1)))}}} = {{{(matrix(1,1,1/(2-1)))}}} = {{{matrix(1,1,1/1))}}} = 1

So the limit exists at 2 and equals 1. However f(2) is not defined.

So there is a removable discontinuity at x=2 

Sometimes this is called "a hole in the graph".

So there is a non-removable infinite discontinuity at x=1,
and a removable discontinuity at x=2`


Edwin</pre>