Let y=f(x)
Find the limits of the function as it tends to the limits of its domain except infinity. If their result is infinity, than that x line is an asymptote. Here, the domain is:
x in(-oo,1)uu(1,3)uu(3,+oo)
So the 4 possible vertical asymptotes are:
lim_(x->1^-)f(x)
lim_(x->1^+)f(x)
lim_(x->3^-)f(x)
lim_(x->3^+)f(x)
Asymptote x->1^-
lim_(x->1^-)f(x)=lim_(x->1^-)(x+1)^2/((x-1)(x-3))=2^2/(0^-*(-2))=
=-2^2/(0*(-2))=4/(0*2)=4/0=+oo Vertical asymptote for x=1
Note: for x-1 since x is slightly lower than 1 the result will be something a bit lower than 0, so the sign will be negative, hence the note 0^- which later translates to a negative sign.
Confirmation for asymptote x->1^+
lim_(x->1^+)f(x)=lim_(x->1^+)(x+1)^2/((x-1)(x-3))=2^2/(0^+*(-2))=
=2^2/(0*(-2))=-4/(0*2)=-4/0=-oo Confirmed
Asymptote x->3^-
lim_(x->3^-)f(x)=lim_(x->3^-)(x+1)^2/((x-1)(x-3))=3^2/(2*0^-)=
=-3^2/(2*0)=-9/0=-oo Vertical asymptote for x=3
Confirmation for asymptote x->3^+
lim_(x->3^+)f(x)=lim_(x->3^+)(x+1)^2/((x-1)(x-3))=3^2/(2*0^+)=
=3^2/(2*0)=9/0=+oo Confirmed
Find both limits as the function tends to +-oo
Minus infinity x->-oo
lim_(x->-oo)f(x)=lim_(x->-oo)(x+1)^2/((x-1)(x-3))=
=lim_(x->-oo)(x^2+2x+1)/(x^2-4x-3)=lim_(x->-oo)(x^2(1+2/x+1/x^2))/(x^2(1-4/x-3/x^2))=
=lim_(x->-oo)(cancel(x^2)(1+2/x+1/x^2))/(cancel(x^2)(1-4/x-3/x^2))=lim_(x->-oo)(1+2/x+1/x^2)/(1-4/x-3/x^2)=
=(1+0+0)/(1-0-0)=1 Horizontal asymptote for y=1
Plus infinity x->+oo
lim_(x->+oo)f(x)=lim_(x->+oo)(x+1)^2/((x-1)(x-3))=
=lim_(x->+oo)(x^2+2x+1)/(x^2-4x-3)=lim_(x->+oo)(x^2(1+2/x+1/x^2))/(x^2(1-4/x-3/x^2))=
=lim_(x->+oo)(cancel(x^2)(1+2/x+1/x^2))/(cancel(x^2)(1-4/x-3/x^2))=lim_(x->+oo)(1+2/x+1/x^2)/(1-4/x-3/x^2)=
=(1+0+0)/(1-0-0)=1 Horizontal asymptote for y=1
Note: it just so happens that this function has a common horizontal for both -oo and +oo. You should always check both.
You must first find both limits:
lim_(x->+-oo)f(x)/x
For each, if this limit is a real number, then the asymptote exists and the limit is its slope. The y intercept of each is the limit:
lim_(x->+-oo)(f(x)-m*x)
However, to save us the trouble, you can use some function "knowledge" to avoid this. Since we know f(x) has horizontal asymptote for both +-oo the only way of having an oblique is having another line as x->+-oo. However, f(x) is a 1-1 function so there can't be two y values for one x, hence a second line is impossible, so it's impossible to have oblique asymptotes.
