There are two vertical asymptotes (#x=-3# and #x=3#) and one horizontal asymptote (#y=0#).
We have #f(x) = (x^2-1)/(x^4-81)=((x+1)(x-1))/((x^2+9)(x+3)(x-3))#.
Therefore, the domain of #f(x)# is :
#D = ]-oo;-3[ uu ]-3;3[ uu ]3;+oo[#.
In order to find the vertical asymptotes, you need to examine
#lim_(x->a)|f(x)|#, where #a# is a forbidden #x#-value of #f(x)#.
(In general, you'll take forbidden #x#-values from the edges of the domain, excluding #+oo# and #-oo#).
Here, the forbidden values at the edges of the domain are #a_1 = -3# and #a_2 = 3#.
There is a vertical asymptote in #x=a# if #lim_(x->a)|f(x)|=+oo#.
#lim_(x->-3)|f(x)|=((-2)*(-4))/(18*0*(-6))=8/0 = +oo#.
#lim_(x->+3)|f(x)|=(4*2)/(18*6*0)=8/0 = +oo#.
Therefore, there are two vertical asymptotes,
one in #x = -3# and another in #x = 3#.
In order to find the horizontal asymptotes, you need to examine
#lim_(x->-oo)f(x)# and #lim_(x->+oo)f(x)#.
There is an horizontal asymptote in #y=h# on the left of your graph/function if #lim_(x->-oo)f(x)=h#.
#lim_(x->-oo)f(x)=lim_(x->-oo)(x^2-1)/(x^4-81) = lim_(x->-oo)x^2/x^4 = lim_(x->-oo)1/x^2 = 1/(+oo) = 0#.
And there is an horizontal asymptote in #y=h# on the right of your graph/function if #lim_(x->+oo)f(x)=h#.
#lim_(x->+oo)f(x)=lim_(x->+oo)(x^2-1)/(x^4-81) = lim_(x->+oo)x^2/x^4 = lim_(x->+oo)1/x^2 = 1/(+oo) = 0#.
Therefore, there is one horizontal asymptote when #y=0#, on both sides of the function.
Since there are already horizontal asymptotes on the left and on the right of the function, there won't be any oblique asymptotes on both sides of it.
That's it.
