y=(x^2 + 2)/ (3x^2 - 5)

Is the asymptote at

x= sqrt. 5/3
y=o

1 answer

vertical asymptote is ok
horizontal asymptote is at y=1/3

(x^2+2)/(3x^2-5)
= (1 + 2/x^2) / (3 - 5/x^2)

as x gets huge, the fractions vanish, and we are left with 1/3

When the degree of top and bottom are the same, then the limit is the ratio of the coefficients of the highest power.
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