AP Calculus AB and BC

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

Q: Why is this 1/3 when (x - sin x)/x^3 is 1/6?

The cubic coefficients differ. Sine has -\frac{x^{3}}{6}, so x - \sin x leaves \frac{x^{3}}{6}, while tangent has +\frac{x^{3}}{3}, so \tan x - x leaves \frac{x^{3}}{3}. Dividing by x^{3} leaves those coefficients as the two answers.

Q: Is there a way to avoid the third pass?

Yes. After one pass, rewrite \sec^{2}x-1 as \tan^{2}x and the expression becomes \frac{1}{3}\left(\frac{\tan x}{x}\right)^{2}, which goes to \frac{1}{3}. The three-pass version is worth seeing once, but the identity is faster and less error-prone under time pressure.

Q: Does the limit exist from both sides?

Yes, and they agree. Replacing x by -x flips the sign of \tan x - x and of x^{3}, so the quotient is unchanged and the graph is symmetric about the vertical axis near 0.

The limit of (tan x - x)/x^3 as x approaches 0 is 1/3, about 0.3333. Direct substitution gives 0 over 0 three stages running, so L'Hopital's rule applies three times. The tangent series has plus x cubed over 3, which is why the answer is 1/3 and not the 1/6 of the sine version.

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

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

Three passes through the rule

The derivative of $\tan x$ is $\sec^{2}x$ , and $\sec^{2}0 = 1$ , so the first pass produces another $\frac{0}{0}$ .

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

Differentiating again uses $\frac{d}{dx}\sec^{2}x = 2\sec^{2}x\tan x$ , and the result simplifies before the next check.

\lim_{x \to 0}\frac{2\sec^{2}x\tan x}{6x} = \lim_{x \to 0}\frac{\sec^{2}x\tan x}{3x}

That is $\frac{1 \cdot 0}{0} = \frac{0}{0}$ , so one more pass is allowed. The product rule handles the numerator.

\lim_{x \to 0}\frac{2\sec^{2}x\tan^{2}x+\sec^{4}x}{3} = \frac{0+1}{3} = \frac{1}{3}

One pass is enough if you know the identity

The Pythagorean identity turns the first result into something familiar: $\sec^{2}x - 1 = \tan^{2}x$ , so $\frac{\sec^{2}x-1}{3x^{2}} = \frac{1}{3}\left(\frac{\tan x}{x}\right)^{2}$ . Since $\frac{\tan x}{x} \to 1$ , the value is $\frac{1}{3}$ after a single differentiation.

The series makes the 3 obvious

The Maclaurin expansion of tangent starts one term past the linear piece.

\tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \cdots

Subtracting $x$ removes the linear term, and the leading survivor is cubic, exactly matching the denominator.

\frac{\tan x-x}{x^{3}} = \frac{1}{3} + \frac{2x^{2}}{15} + \cdots \;\longrightarrow\; \frac{1}{3}

The correction term is positive here, so the values come down to $\frac{1}{3}$ from above, the mirror image of the sine version, which climbs to $\frac{1}{6}$ from below.

Why direct substitution fails

With $\tan 0 = 0$ , the numerator and the denominator vanish together.

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

Replacing $\tan x$ by $x$ for small angles does not rescue it, for the same reason as in the sine case: the numerator is the error in that approximation, and the error is of size $\frac{x^{3}}{3}$ , the same order as the denominator. Discarding it discards the answer.

$x$	$\frac{\tan x-x}{x^{3}}$
0.1	0.3346721
0.01	0.3333467
0.001	0.3333335

The function is even, so the left-hand values repeat these, and the two-sided limit exists. Stay well inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ when sampling, since tangent blows up at the ends of that interval.

The mistakes students make

Answering $\frac{1}{6}$ by pattern-matching the sine version. Same shape, different series coefficient, and the answers differ by a factor of 2.
Differentiating $\tan x$ as $\sec x\tan x$ . That is the derivative of $\sec x$ ; tangent gives $\sec^{2}x$ .
Reporting a negative answer. For $0 < x < \frac{\pi}{2}$ the ordering is $\sin x < x < \tan x$ , so $\tan x - x$ is positive, and the even symmetry keeps the quotient positive on the left as well.
Applying the quotient rule to the whole fraction instead of differentiating the numerator and denominator separately.
Stopping at $\frac{\sec^{2}x-1}{3x^{2}}$ and calling it 0. The numerator does vanish, but so does the denominator, so the form is unresolved.

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 is this 1/3 when (x - sin x)/x^3 is 1/6?

The cubic coefficients differ. Sine has $-\frac{x^{3}}{6}$ , so $x - \sin x$ leaves $\frac{x^{3}}{6}$ , while tangent has $+\frac{x^{3}}{3}$ , so $\tan x - x$ leaves $\frac{x^{3}}{3}$ . Dividing by $x^{3}$ leaves those coefficients as the two answers.

Is there a way to avoid the third pass?

Yes. After one pass, rewrite $\sec^{2}x-1$ as $\tan^{2}x$ and the expression becomes $\frac{1}{3}\left(\frac{\tan x}{x}\right)^{2}$ , which goes to $\frac{1}{3}$ . The three-pass version is worth seeing once, but the identity is faster and less error-prone under time pressure.

Does the limit exist from both sides?

Yes, and they agree. Replacing $x$ by $-x$ flips the sign of $\tan x - x$ and of $x^{3}$ , so the quotient is unchanged and the graph is symmetric about the vertical axis near 0.