If the numerator degree is less than denominator degree, what is the oblique or slant asymptote?
T(x) = (x^3)/(x^4-1)
This answer on StackExchange may help.
There is NO "oblique or slant" asymptote. An "asymptote" is a line or curve that a function graph approaches as x grows large, either positive or negative. An "oblique or slant" is a non vertical or horizontal line the function graph approaches and that requires that the function be of the form f(x)= mx+ b+ F(x) where F(x) goes to 0 as x gets larger. That can only happen if the numerator has degree exactly one more than the denominator (and other curves as asymptotes if the degree of the numerator is two or more higher). As long as the degree of the denominator is greater than the degree of the numerator, the horizontal line, y= 0, is the asymptote.
To add about the relevance of the link: As post #4 says, f(x) has an asumptote mx+b if . The link describes how to compute the limit of a rational function f(x): one just needs to consider the ratio of the senior terms in the numerator and the denominator. Here
so the asimptote is y = 0.
Thank you everyone. Say find the asymptotes for (4x)/(x^2-49).
The vertical asymptotes are found by equating the denominator to zero and solving for x.
Since the degree of the top is smaller than denominator degree, the horizontal asymptote is the line y = 0 which is
the x-axis, right?
There is no slant asymptote here, right?