AP Calculus AB and BC

Limit of (sin x - x)/x as x Approaches 0 Is 0

Q: What if the denominator is x^3?

Then the limit is -\frac{1}{6}, because \sin x - x vanishes to third order.

The limit of sin x minus x, over x, as x approaches 0 is 0. Splitting the fraction gives sin x over x, minus 1, and since the first piece tends to 1 the difference tends to 0.

\lim_{x \to 0} \frac{\sin x - x}{x} = 0

Settled by splitting the fraction into a known limit.

Split before you differentiate

\frac{\sin x - x}{x} = \frac{\sin x}{x} - 1 \longrightarrow 1 - 1 = 0

L'Hopital works too and gives $\cos x - 1 \to 0$ , but splitting reuses a limit you already know and takes one line.

The denominator power decides everything

$\sin x - x$ vanishes to THIRD order, since the series is $-\frac{x^{3}}{6} + \cdots$ . So dividing by $x$ leaves something still heading to $0$ , while dividing by $x^{3}$ gives the finite value $-\frac{1}{6}$ .

\frac{\sin x - x}{x} \to 0, \qquad \frac{\sin x - x}{x^{3}} \to -\frac{1}{6}

The mistakes students make

Answering $1$ by treating the whole thing as the special limit.
Cancelling the $x$ terms. The numerator is a difference, not a product.

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)/x as x approaches 0?

It is $0$ .

What if the denominator is x^3?

Then the limit is $-\frac{1}{6}$ , because $\sin x - x$ vanishes to third order.