AP Calculus AB and BC

Limit of x^3 as x Approaches Negative Infinity

The limit of x cubed as x approaches negative infinity is negative infinity. Cubing keeps the sign of the input, so large negative inputs give large negative outputs. That is the difference from x squared, which runs to positive infinity at both ends.

lim⁡x→−∞x3=−∞\lim_{x \to -\infty} x^3 = -\infty

Settled by end behaviour of an odd power.

Why the sign survives the cube

A product of three negative factors is negative: two of them cancel each other and one is left over. So x3x^{3} carries the same sign as xx at every input, while the magnitude climbs fast, since (−100)3=−106(-100)^{3} = -10^{6}.

xxx3x^{3}
−10-10−103-10^{3}
−100-100−106-10^{6}
−103-10^{3}−109-10^{9}
−106-10^{6}−1018-10^{18}

There is no floor for the values to settle on. For any bound M>0M > 0, every x<−M3x < -\sqrt[3]{M} gives x3<−Mx^{3} < -M, so the outputs eventually pass below whatever level you set.

lim⁡x→−∞x3=−∞\lim_{x \to -\infty} x^{3} = -\infty

Why substitution fails, and why no form appears

−∞-\infty is not an input. Writing (−∞)3(-\infty)^{3} describes the process rather than performing a calculation, and it earns nothing as a written step.

Nothing indeterminate turns up here either. Indeterminate forms need two competing behaviours, and a single power has none. Two facts decide the answer: the parity of the exponent, and the direction of the approach.

FunctionAt −∞-\inftyAt ∞\infty
x2x^{2}∞\infty∞\infty
x3x^{3}−∞-\infty∞\infty
−x3-x^{3}∞\infty−∞-\infty
x4x^{4}∞\infty∞\infty

The same reading extends to any polynomial, because the leading term outgrows everything under it. x3+500x2x^{3} + 500x^{2} is still positive at x=−100x = -100, yet its limit at −∞-\infty is −∞-\infty, since the cube takes charge once ∣x∣|x| passes 500.

Odd degree forces a crossing

A polynomial of odd degree has opposite infinite limits at its two ends, so its values sweep from below any level to above any level. The Intermediate Value Theorem then guarantees a real root, which is why every cubic crosses the xx-axis while some quartics never do.

The mistake students make

  • Answering ∞\infty on the reasoning that raising to a power makes things positive. Only even powers do that. Odd powers hand back whatever sign came in.
  • Writing that the limit does not exist and stopping. It is not a real number, true, but −∞-\infty names the behaviour, and that is what the rubric asks for.
  • Mixing up −x3-x^{3} and (−x)3(-x)^{3}. They are the same function, and both run to ∞\infty as x→−∞x \to -\infty, which is the opposite of the plain cube.
  • Checking one end only. x3x^{3} has different infinite limits on the two sides, so an end behaviour question needs both of them.

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

Why does x squared behave differently?

An even power discards the sign of the input: (−1000)2=106(-1000)^{2} = 10^{6}, exactly as 100021000^{2} does. So x2x^{2} runs to ∞\infty at both ends while x3x^{3} splits, and in general the parity of the exponent decides whether the two ends agree.

Is negative infinity an acceptable answer?

Yes, and it is the expected one, because it records that the values fall without bound. Topic 1.15 pairs limits at infinity with horizontal asymptotes, and an infinite limit at an infinite end means there is no horizontal asymptote on that side.

How do I handle a full polynomial quickly?

Keep the leading term and drop the rest. For 4x5−9x3+24x^{5} - 9x^{3} + 2 the limit at −∞-\infty matches that of 4x54x^{5}, which is −∞-\infty. Only the parity of the degree and the sign of the leading coefficient matter, and a negative coefficient flips the answer.