How do you find all the asymptotes for function #f(x)=((x^2)-1)/((x^4)-81)#?

Answer 1
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.

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Answer 2

To find the asymptotes of the function ( f(x) = \frac{x^2 - 1}{x^4 - 81} ), we need to consider both vertical and horizontal asymptotes.

Vertical Asymptotes: Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. To find the vertical asymptotes, we set the denominator equal to zero and solve for ( x ):

[ x^4 - 81 = 0 ]

[ (x^2 + 9)(x^2 - 9) = 0 ]

[ (x^2 + 9)(x + 3)(x - 3) = 0 ]

This gives us three potential vertical asymptotes at ( x = -3 ), ( x = 3 ), and ( x = -\sqrt{9} ), which is the same as ( x = -3 ), and ( x = \sqrt{9} ), which is the same as ( x = 3 ).

Horizontal Asymptotes: Horizontal asymptotes occur when ( x ) approaches positive or negative infinity. To find the horizontal asymptotes, we examine the degree of the numerator and the denominator of the function.

The degree of the numerator is 2, and the degree of the denominator is 4. When the degree of the numerator is less than the degree of the denominator, as in this case, there is a horizontal asymptote at ( y = 0 ) (the x-axis).

Therefore, the horizontal asymptote for ( f(x) ) is ( y = 0 ).

In summary:

  • Vertical asymptotes occur at ( x = -3 ) and ( x = 3 ).
  • The horizontal asymptote occurs at ( y = 0 ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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