Find the vertical asymptotes. For each vertical asymptote x=a,...

Question

Find the vertical asymptotes. For each vertical asymptote x=a, analyze lim x→a^- f(x) and lim x→a⁺ f(x).

f(x) = (x^4−1)/(x^2−1)

Accepted Answer

Welcome back, everyone. Identify any vertical asymptotes of the function F of X equals X to the power of 4 minus 625 divided by X2 minus 25. Evaluate the limits of the function as X approaches A from the left and from the right for each vertical asymptote in a form X equals A. So in this problem we're given a rational function, right? It consists of a numerator and a denominator. We have x the power of 4 minus 625 in the numerator. In the denominator we have X2 minus 25. We have to recall that if we have vertical asymptotes, well, essentially they make the denominator equal to 0. The function does not exist at the denominator of 0. So essentially what we're going to do is basically set X2 minus 25, which is our denominator. Equal to 0. However, we have to be extra careful. If our solutions correspond to removable discontinuities such as holes, then such x values do not represent vertical asymptotes. So what we want to do is basically perform a complete factorization of the denominator. Let's understand that X2 minus 25 is a difference of squares. X2 minus 5 squad, which can be factored out as X minus 5 multiplied by X + 5. Also, let's notice that our numerator is X minus. 625 which is X2 squad minus 25 squad. Once again we have a difference of squares, and this can be factored out as x2 minus 25 multiplied by X2 + 25. And notice that we end up with the same X2 minus 25. So X2 minus 25 can be rewritten as X minus 5 multiplied by X + 5. And we can rewrite F of X in its completely factored form. So that would be X minus 5 in the numerator, multiplied by X + 5. Multiplied by x2 + 25 and in the denominator, we have X minus 5 multiplied by X + 5. Now if we set our denominator equal to 0 in a factored form, we can clearly see that there are two solutions x equals positive or -5, and those two are the potential vertical asymptotes. But in a factored form we notice. That each factor cancels out. X minus 5 can be canceled out because there's x minus 5 in the numerator and also X + 5 can be canceled out because we have X + 5 in the numerator. So X equals -5. And x equals 5. Those are removable discontinuities. They are holes. If they weren't holes, we could then say that we had two vertical asymptotes, but in this case, well, this is not true. Those are holes, and as a result, we don't have any vertical asymptotes since there are no more factors to consider in the denominator. So the final answer to this problem is no vertical asymptotes. That's our final answer and thank you for watching.

Find the vertical asymptotes. For each vertical asymptote x=a, analyze lim x→a^- f(x) and lim x→a⁺ f(x).
f(x) = (x^4−1)/(x^2−1)

Key Concepts

Vertical Asymptotes

Limits

Factoring Polynomials

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