AP Calculus AB and BC

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

Q: What is the general rule?

Near 0 both \sin(ax) and \tan(ax) behave like ax, so any such ratio tends to the ratio of the coefficients.

The limit of sin 3x over tan 2x as x approaches 0 is three halves. Near 0 both sine and tangent behave like their arguments, so the ratio is 3x over 2x. Writing each factor against its own angle makes the standard limits do the work.

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

Settled by matching each inner angle to the standard limits.

Match each angle to itself

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

Each of the two quotients tends to $1$ , leaving the coefficient ratio $\frac{3}{2}$ .

The shortcut worth knowing

Near 0, sin(ax) and tan(ax) both behave like ax. So any ratio of these functions tends to the ratio of the coefficients, and this whole family can be read off in one step.

L'Hopital agrees

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

The chain rule supplies both coefficients, which is the same $\frac{3}{2}$ the algebraic method pulls out front.

The mistakes students make

Answering $1$ by treating any sine over tangent as the standard limit.
Inverting to $\frac{2}{3}$ . The numerator's coefficient goes on top.

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 3x / tan 2x at 0?

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

What is the general rule?

Near $0$ both $\sin(ax)$ and $\tan(ax)$ behave like $ax$ , so any such ratio tends to the ratio of the coefficients.