How do you find the horizontal asymptote for #(3x^5 + 1) / (2x^6 + 3x -1)#?
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"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
1 Answer
Jun 18, 2018
HA=
Explanation:
if the degree of the x is higher on the bottom than the top, the horizonatl asymptote is 0.
