AP Calculus AB and BC

Limit of (x - sin x)/x^3 as x Approaches 0 Is 1/6

Q: Why does the 6 appear in the answer?

It is the 3! in the series for sine. The cubic term of \sin x is -\frac{x^{3}}{3!} = -\frac{x^{3}}{6}, so subtracting sine from x leaves \frac{x^{3}}{6} as the leading behaviour, and dividing by x^{3} leaves the coefficient standing alone.

The limit of (x - sin x)/x^3 as x approaches 0 is 1/6, about 0.1667. Direct substitution gives 0 over 0, and so do the next two stages, so L'Hopital's rule runs three times. The series route reads it off faster: the cubic term of sin x is minus x cubed over 6.

\lim_{x \to 0} \frac{x-\sin x}{x^3} = \frac{1}{6}

Settled by L'Hopital's rule applied three times.

Three passes, each one checked first

Every pass differentiates the top and the bottom separately, and every pass is legal only because the form was re-checked and came back $\frac{0}{0}$ .

\lim_{x \to 0}\frac{x-\sin x}{x^{3}} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{1-\cos x}{3x^{2}}

At $x = 0$ that is $\frac{1-1}{0} = \frac{0}{0}$ again, so the rule applies a second time.

\lim_{x \to 0}\frac{1-\cos x}{3x^{2}} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{\sin x}{6x}

Still $\frac{0}{0}$ . One more pass leaves a form that substitution can finally handle.

\lim_{x \to 0}\frac{\sin x}{6x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{\cos x}{6} = \frac{1}{6}

Stop one pass early if you like

The third stage $\frac{\sin x}{6x}$ is $\frac{1}{6} \cdot \frac{\sin x}{x}$ , and the special trig limit sends $\frac{\sin x}{x}$ to 1. That gives $\frac{1}{6}$ with two passes instead of three, and it avoids leaning on L'Hopital for a limit that was proved by the squeeze theorem in the first place.

The same answer from the series

The Maclaurin series makes the result look inevitable rather than lucky.

\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots

Subtracting from $x$ cancels the linear term, so the numerator starts at order three, matching the denominator.

\frac{x-\sin x}{x^{3}} = \frac{\frac{x^{3}}{6} - \frac{x^{5}}{120} + \cdots}{x^{3}} = \frac{1}{6} - \frac{x^{2}}{120} + \cdots

Sending $x \to 0$ kills every remaining term and leaves $\frac{1}{6}$ . The correction term is negative, so the values climb toward $\frac{1}{6}$ from below, which matches a table.

Why direct substitution fails

Since $\sin 0 = 0$ , both parts vanish at once.

\frac{0-\sin 0}{0^{3}} = \frac{0}{0}

The tempting repair is the small-angle approximation $\sin x \approx x$ , which makes the numerator look like 0 and suggests an answer of 0. That approximation is too coarse here, because the numerator is precisely the error it throws away, and that error has size $\frac{x^{3}}{6}$ : the same order as the denominator.

$x$	$\frac{x-\sin x}{x^{3}}$
0.1	0.1665834
0.01	0.1666658
0.001	0.1666667

The function is even, since replacing $x$ by $-x$ flips the sign of both the numerator and the denominator, so the left-hand values match these exactly.

The mistakes students make

Quitting after one pass and reporting $\frac{1-\cos x}{3x^{2}}$ as 0 or as undefined. It is still $\frac{0}{0}$ , and its value is still $\frac{1}{6}$ .
Using the quotient rule instead of differentiating top and bottom separately. L'Hopital's rule is not the derivative of the fraction.
Applying a fourth pass to $\frac{\cos x}{6}$ . Once the form is determinate the rule no longer applies, and differentiating anyway gives $\frac{-\sin x}{0}$ , which is meaningless.
Answering $\frac{1}{3}$ , which belongs to the tangent version $\frac{\tan x - x}{x^{3}}$ . The two look alike and have different cubic coefficients.
Replacing $\sin x$ by $x$ inside the numerator to get 0. Approximations may be substituted only where the discarded term is smaller than what survives.

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 the 6 appear in the answer?

It is the $3!$ in the series for sine. The cubic term of $\sin x$ is $-\frac{x^{3}}{3!} = -\frac{x^{3}}{6}$ , so subtracting sine from $x$ leaves $\frac{x^{3}}{6}$ as the leading behaviour, and dividing by $x^{3}$ leaves the coefficient standing alone.

Do I really have to check the form before every pass?

Yes. L'Hopital's rule is a statement about indeterminate quotients, so applying it to a determinate one gives a number with no connection to the limit. Checking is one line: substitute and see whether $\frac{0}{0}$ or $\frac{\infty}{\infty}$ comes back.

Can this be done without L'Hopital's rule at all?

Yes, two ways. The series argument above is complete on its own, and there is an algebraic route using $x - \sin x$ with the half-angle identity, though it is longer. For AP purposes the three-pass version and the series version are the two you are expected to recognise.