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Calculus/horizontal asymptotes...

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Expert: - 2/6/2008


Question
QUESTION: My question says to find the horizontal asymptotes of:

f(x)= sqrt(x^2 + x + 1) - sqrt(x^2-x)


I really dont know where to begin on this one other than set them as inequalities.

ANSWER: The slope where a horizontal asymptote occurs is 0, therefore take the derivative and set it equal to 0.  It will probably take a little bit of work to get a common denominator.  Also note the domain and range of the function and derivative.

---------- FOLLOW-UP ----------

QUESTION: Alright I have [(2x+1)/(2sqrt(x^2+x+1)] - [(2x-1)/(2sqrt(x^2-x)].  Now, what do I do with it?  How do you get the common denominator, you cant multiply by a conjugate, I dont think.

Answer
If you want a common denominator, since the square roots are usually in the numerator, the problem might get a whole lot messier.

What needs to be done is multiply the first one by
sqrt(x^2+x+1)(x^2-x) in the top and bottom and multiply the second by
(x^2+x+1)sqrt(x^2-x).  Both fractions will then have
2(x^2+x+1)(x^2-x) = 2x(x-1)(x^2+x+1) in the denomiator and the numerator may or may not be simplified.

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