How do you find the horizontal asymptote of(2x^2+x-7)/(x^2-1)?

Question

horizontal asymptote of(2x^2+x-7)/(x^2-1)?

Answer 1

lim[x ~> 0] (2x^2 + x - 7)/(x^2 - 1)
lim[x ~> 0] (4x + 1)/(2x) L'Hopital
lim[x ~> 0] 4/2 L'Hopital
2
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lim[x ~> 0] (2x^2 + x - 7)/(x^2 - 1)
lim[x ~> 0] (2x^2/x^2 + x/x^2 - 7/x^2)/(x^2/x^2 - 1/x^2)
lim[x ~> 0] (2 + 1/x - 7/x^2)/(1 - 1/x^2)
lim[x ~> 0] (2 + 0 - 0)/(1 - 0) = 2

Answer 2

Find the horizontal asymptote.

(2x² + x - 7)/(x² - 1)

Divide numerator and denominator by x² and take the limit
as x→∞.

Limit as x→∞ {(2 + 1/x - 7/x²)/(1 - 1/x²)} = 2/1 = 2

All the other terms go to zero.

Answer 3

Take the lim as x ---> infinity

lim = 2 = horizontal asymptote

Answer 4

Divide each and every term by potential of e^x, then = a million / (e^-x + a million) be conscious, the shrink as x->+? of e^-x = lim (a million/e^x) = 0 So the shrink of the completed expression is a million / (0+a million) = a million. however the shrink as x->-? of e^-x = lim (a million/e^x) = ? So the shrink of the completed expression is a million / ? = 0.