AP Calculus AB and BC

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

Q: How is it different from (tan x - x)/x^3?

That one converges to \frac{1}{3}, because subtracting x removes the leading term.

The limit of tan x over x cubed as x approaches 0 is infinity. Split it as tan x over x, times 1 over x squared. The first factor tends to 1 and the second grows without bound, positively from both sides because of the square.

\lim_{x \to 0} \frac{\tan x}{x^{3}} = \infty

Settled by peeling off the standard trigonometric limit.

Split off what you know

\frac{\tan x}{x^{3}} = \frac{\tan x}{x}\cdot\frac{1}{x^{2}} \longrightarrow 1 \cdot (+\infty) = \infty

The leftover power is even, so $\frac{1}{x^{2}}$ is positive on both sides and the two one-sided limits agree.

Compare with the sine version

$\frac{\sin x}{x^{3}}$ behaves identically, since $\sin x$ and $\tan x$ agree to first order near $0$ . But $\frac{\tan x - x}{x^{3}}$ is completely different, converging to $\frac{1}{3}$ , because subtracting $x$ cancels the leading behaviour.

The mistakes students make

Answering $1$ by treating the whole thing as the standard limit. Only one factor of $x$ is consumed.
Assuming the two sides disagree. An even leftover power keeps the sign the same.

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 tan x / x^3 at 0?

It is $\infty$ , from both sides.

How is it different from (tan x - x)/x^3?

That one converges to $\frac{1}{3}$ , because subtracting $x$ removes the leading term.