How do you find the asymptotes for #(x^2+1)/(x^2-1)#?

Answer 1

The vertical asymptotes are #x=1# and #x=-1#
The horizontal asymptote is #y=1#

The denominator cannot be divided by #0#. So the vertical asymptotes are #x=1# and #x=-1# As the degree of the numerator and denominator are the same, we would not expect an oblique asymptote. Limit #(x^2-1)/(x^2-1)=x^2/x^2=1# #x->oo# graph{(y-(x^2+1)/(x^2-1))(y-1)=0 [-7.9, 7.9, -3.95, 3.95]}

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Answer 2

Benjamin Achtenberg

To find the asymptotes for ( \frac{x^2 + 1}{x^2 - 1} ), we first look for vertical asymptotes by setting the denominator equal to zero and solving for ( x ).

( x^2 - 1 = 0 )

( (x - 1)(x + 1) = 0 )

( x = 1 ) or ( x = -1 )

So, there are vertical asymptotes at ( x = 1 ) and ( x = -1 ).

Next, we check for horizontal asymptotes. Since the degrees of the numerator and denominator are equal, we divide the leading coefficients to find the horizontal asymptote.

The leading coefficients are both 1.

Therefore, the horizontal asymptote is ( y = \frac{1}{1} = 1 ).

So, the asymptotes for ( \frac{x^2 + 1}{x^2 - 1} ) are ( x = 1 ), ( x = -1 ), and ( y = 1 ).

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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.