AP Calculus AB and BC

Limit of sin x / x^3 at 0 Is Infinity

The limit of sin x over x cubed as x approaches 0 is infinity. Split it as sin x over x, times 1 over x squared. The first factor tends to 1 and the second grows without bound, and because x squared is positive on both sides the limit is plus infinity from each direction.

lim⁡x→0sin⁡xx3=∞\lim_{x \to 0} \frac{\sin x}{x^{3}} = \infty

Settled by peeling off the special trigonometric limit.

Separating the known part

Split off the factor you already know, and look at what is left.

sin⁡xx3=sin⁡xx⋅1x2\frac{\sin x}{x^{3}} = \frac{\sin x}{x}\cdot\frac{1}{x^{2}}

The first factor tends to 11. The second is 1x2\frac{1}{x^{2}}, which grows without bound, and crucially it is POSITIVE on both sides of zero because the square kills the sign.

lim⁡x→0sin⁡xx3=1⋅(+∞)=∞\lim_{x \to 0}\frac{\sin x}{x^{3}} = 1 \cdot \left(+\infty\right) = \infty

Why the odd power would differ

With sin x over x to the fourth the same argument gives 1 over x cubed, which is positive on the right and negative on the left, so that limit does not exist. The parity of the leftover power decides it.

What saying infinity actually claims

Writing that a limit equals infinity is a statement that the limit FAILS to exist, in a specific and describable way: the values exceed any bound you name. On free response, saying the limit does not exist and then describing the unbounded behaviour is the answer that earns full credit.

The graph has a vertical asymptote at x=0x = 0, and the function rises on both sides of it.

The mistakes students make

  • Answering 11 by treating sin⁡xx3\frac{\sin x}{x^{3}} as the special limit. Only one factor of xx is consumed by it.
  • Answering that the limit does not exist without describing the behaviour. Here both sides agree on +∞+\infty, which is a much more informative answer.
  • Assuming any leftover power gives infinity from both sides. An odd leftover power changes sign across zero, and then the two sides disagree.

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

What is the limit of sin x / x^3 as x approaches 0?

It is ∞\infty, meaning the function grows without bound from both sides.

Why is it the same from both sides?

The leftover factor after using the special limit is 1x2\frac{1}{x^{2}}, which is positive whichever side you approach from.

What about sin x / x^4?

That limit does not exist. The leftover 1x3\frac{1}{x^{3}} is positive on the right and negative on the left, so the two sides disagree.