AP Calculus AB and BC

Limit of tan 2x / tan 3x at 0 Is 2/3

Q: What is the general rule for tan(ax) over tan(bx)?

It tends to \frac{a}{b} as x \to 0, because each tangent behaves like its own angle and the x cancels.

Q: Does L'Hopital's rule work here?

Yes. Differentiating gives \frac{2\sec^{2}2x}{3\sec^{2}3x}, and at x = 0 both secants equal 1, leaving \frac{2}{3}.

The limit of tan 2x over tan 3x as x approaches 0 is two thirds. Near 0 each tangent behaves like its own angle, so the quotient behaves like 2x over 3x. The x cancels and the coefficients alone decide the value.

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

Settled by matching each inner angle to the standard limits.

Give each tangent its own angle

The standard limit $\lim_{u \to 0}\frac{\tan u}{u} = 1$ needs the angle underneath to match the angle inside. Build both pairs and put the leftover constants out front.

\frac{\tan 2x}{\tan 3x} = \frac{2}{3}\cdot\frac{\tan 2x}{2x}\cdot\frac{3x}{\tan 3x}

Both quotients tend to $1$ as $x \to 0$ , since $2x$ and $3x$ each tend to $0$ along with $x$ . What survives is the coefficient ratio $\frac{2}{3}$ .

The rule this problem is really testing

For nonzero constants a and b, the ratio of tan ax to tan bx tends to a over b as x approaches 0. Sine may replace either tangent without changing anything, since sin u and tan u both behave like u near 0.

Where the tangent rule comes from

Tangent inherits its behaviour from sine. Writing $\tan u = \frac{\sin u}{\cos u}$ splits the standard limit into two familiar pieces.

\frac{\tan u}{u} = \frac{\sin u}{u}\cdot\frac{1}{\cos u} \longrightarrow 1 \cdot 1 = 1

Numbers agree. At $x = 0.01$ the quotient $\frac{\tan 0.02}{\tan 0.03}$ is about $0.6666$ , already within a thousandth of $\frac{2}{3}$ .

The mistakes students make

Careless coefficients explain the first two errors below. The third is different: it reads an indeterminate form as a value.

Inverting the answer to $\frac{3}{2}$ . The coefficient from the numerator belongs on top.
Answering $1$ because both tangents vanish at $0$ . They vanish at rates set by their coefficients, and those rates are what the limit compares.
Substituting $x = 0$ , reading $\frac{0}{0}$ , and answering $0$ . That form carries no value of its own.

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 2x / tan 3x as x approaches 0?

It is $\frac{2}{3}$ .

What is the general rule for tan(ax) over tan(bx)?

It tends to $\frac{a}{b}$ as $x \to 0$ , because each tangent behaves like its own angle and the $x$ cancels.

Does L'Hopital's rule work here?

Yes. Differentiating gives $\frac{2\sec^{2}2x}{3\sec^{2}3x}$ , and at $x = 0$ both secants equal $1$ , leaving $\frac{2}{3}$ .