In a non-piece-wise function like this one, a function is discontinuous when the #f(x)# is undefined for a given value of #x#.
Now, assuming you meant #f(x)=x/x^2-x#, we see that the function is undefined when #x^2=0#.
=>#x^2=0#
=>#x=+-sqrt0#
=>#x=0# We have an asymptote #x=0#
Now, assuming you meant #f(x)=x/(x^2-x)#, we see that the function is undefined when #x^2-x=0#.
=>#x^2-x=0#
=>#x(x-1)=0#
=>#1=x=0#
Now, since the numerator #(x+0)# and the denominator #(x+0)(x-1)# share #x+0# in common, we see that we have an asymptote #x=1#
When #x=0#, we find the value we have skipped by solving #lim_(x->0)(x/(x^2-x))#
Since we get #0/0# after plugging 0 in the place of #x#, we simplify this:
#x/(x^2-x)=>cancelx/(cancelx(x-1))=>1/(x-1)#
Therefore,
#lim_(x->0)(x/(x^2-x))=1/(0-1)#
=>#lim_(x->0)(x/(x^2-x))=1/-1#
=>#lim_(x->0)(x/(x^2-x))=-1#
We have a hole at #(0,-1)#
Now, assuming you meant #f(x)=x/(x^(2-x))#, we see that the function is undefined when #x^(2-x)=0#.
For any real number of #x#, the only way to make the equation true is when the base #x# is equal to 0.
We will not have a hole discontinuity here.
