What are the oblique asymptote of #P(x)= 4x^5/(x^3-1)#?

Question

Liam Brennan · Accepted Answer

The function has no oblique (linear) asymptotes.

When we do the division, we get: #P(x)= (4x^5)/(x^3-1) = 4x^2+(4x^2)/(x^3-1)# Although #P(x)# does get arbitrarily close to the parabola #y=4x^2#, a parabola is not an oblique asymptote.

Elias Ackerman · Answer

undefined To find the oblique asymptote of the function P(x) = 4x^5 / (x^3 - 1), we first perform polynomial long division to divide the numerator by the denominator. After performing the division, the quotient will represent the linear part of the oblique asymptote. The remainder will indicate any remaining terms beyond the linear part.

The long division yields:

4x^2 + 0x + 4

Therefore, the quotient of the division is 4x^2 + 4. This represents the linear part of the oblique asymptote.

To find the remainder, we subtract the product of the divisor (x^3 - 1) and the quotient (4x^2 + 4) from the original function:

P(x) - (x^3 - 1)(4x^2 + 4)

After simplifying, the remainder turns out to be 4x + 4.

Therefore, the oblique asymptote of the function P(x) is given by the linear function y = 4x^2 + 4, and there is an additional term 4x + 4 which represents the remainder but does not contribute to the oblique asymptote.