AP Calculus AB and BC

Limit of arcsin(3x)/x as x Approaches 0

Q: Does the same hold for arctangent?

Yes. \frac{\arctan kx}{x} \to k by the identical argument, since arctangent also behaves like its input near zero.

Q: Is there a domain restriction to worry about?

Arcsine needs its input in [-1, 1], so 3x must stay there. Since x \to 0, that is automatic for x near enough to zero.

The limit of arcsine of 3x over x as x approaches zero is three. Inverse sine behaves like its input near zero, exactly as sine does, so matching the denominator to 3x leaves a factor of three.

\lim_{x \to 0} \frac{\arcsin 3x}{x} = 3

Settled by matching the inner expression to the denominator.

The inverse obeys the same rule

Substituting $u = \arcsin(3x)$ , so that $3x = \sin u$ and $u \to 0$ , turns the standard limit into its own reciprocal:

\lim_{u \to 0}\frac{u}{\sin u} = 1 \quad\Longrightarrow\quad \lim_{x \to 0}\frac{\arcsin x}{x} = 1

So $\frac{\arcsin 3x}{x} = 3 \cdot \frac{\arcsin 3x}{3x} \to 3$ . The coefficient survives exactly as it does for sine.

A useful family

Near zero, $\sin u$ , $\tan u$ , $\arcsin u$ , $\arctan u$ and $\ln(1+u)$ all behave like $u$ itself. Any quotient of two of them with linear inner expressions collapses to the ratio of the coefficients.

Recognising the family saves several lines over L'Hopital's rule and never needs the indeterminate form re-checked between steps.

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

Does the same hold for arctangent?

Yes. $\frac{\arctan kx}{x} \to k$ by the identical argument, since arctangent also behaves like its input near zero.

Is there a domain restriction to worry about?

Arcsine needs its input in $[-1, 1]$ , so $3x$ must stay there. Since $x \to 0$ , that is automatic for x near enough to zero.