AP Calculus AB and BC

Limit of ln(1+3x)/x as x Approaches 0

The limit of the natural log of one plus 3x over x as x approaches zero is three. Near zero the logarithm of one plus a small quantity behaves like that quantity itself, so only the coefficient three survives.

limx0ln(1+3x)x=3\lim_{x \to 0} \frac{\ln(1+3x)}{x} = 3

Settled by substitution to a standard logarithm limit.

Match the denominator

The standard limit is limu0ln(1+u)u=1\lim_{u \to 0}\frac{\ln(1+u)}{u} = 1. Here u=3xu = 3x, so the denominator wants to be 3x3x.

ln(1+3x)x=3ln(1+3x)3x31=3\frac{\ln(1+3x)}{x} = 3 \cdot \frac{\ln(1+3x)}{3x} \longrightarrow 3 \cdot 1 = 3

Same discipline as the trigonometric versions: make the denominator match what is inside, and pay for it with a constant.

Where the standard limit comes from

It is the derivative of ln(1+u)\ln(1+u) at u=0u = 0, which is 11+u\frac{1}{1+u} evaluated at 0, namely 1. Equivalently, the Maclaurin series is ln(1+u)=uu22+\ln(1+u) = u - \frac{u^{2}}{2} + \cdots, whose leading term is uu.

The series view also explains the size estimate: ln(1+3x)3x\ln(1+3x) \approx 3x for small xx, so the quotient is approximately 3 before any limit is taken.

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 general rule?

limx0ln(1+kx)x=k\lim_{x \to 0}\frac{\ln(1+kx)}{x} = k for any constant k. Match the denominator to the inner expression and the coefficient falls out.

Why must it be ln(1+3x) rather than ln(3x)?

Because ln(3x)\ln(3x) runs to -\infty as x0x \to 0, so there is no indeterminate form and no finite limit. The 1 inside is what keeps the logarithm near zero.