AP Calculus AB and BC

Limit of x^3/(x^2+1) as x Approaches Infinity

The limit of x cubed over x squared plus one as x approaches infinity is infinity. The numerator has the higher degree, so the quotient grows without bound rather than settling, and the graph has a slant behaviour instead of a horizontal asymptote.

\lim_{x \to \infty} \frac{x^{3}}{x^{2}+1} = \infty

Settled by comparing leading degrees.

Degree decides it

For a quotient of polynomials at infinity there are three cases, decided by comparing degrees. Numerator smaller gives 0; equal gives the ratio of leading coefficients; numerator larger gives an unbounded limit.

\frac{x^{3}}{x^{2}+1} = \frac{x}{1 + 1/x^{2}} \longrightarrow \infty

Dividing top and bottom by $x^{2}$ , the dominant power in the denominator, makes it explicit: what is left is essentially $x$ .

No horizontal asymptote, but structure anyway

Since the degree gap is exactly one, polynomial division gives $\frac{x^{3}}{x^{2}+1} = x - \frac{x}{x^{2}+1}$ , and the subtracted piece tends to 0. So the graph approaches the LINE $y = x$ .

That line is a slant asymptote. It exists precisely when the numerator's degree is one more than the denominator's; a gap of two or more gives no linear asymptote at all.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

When does a rational function have a slant asymptote?

When the numerator's degree is exactly one more than the denominator's. Do the polynomial division; the quotient is the line and the remainder term vanishes.

Can a graph have both a horizontal and a slant asymptote?

Not on the same side. It can have a horizontal asymptote one way and a slant the other, but the two behaviours cannot happen at the same end.