AP Calculus AB and BC

Limit of ln(1 + 2x) / x as x Approaches 0 Is 2

Q: What is the general rule?

\lim_{x \to 0}\frac{\ln(1+ax)}{x} = a for any constant a.

The limit of the natural log of 1 plus 2x, divided by x, as x approaches 0 is 2. Rewrite it as 2 times ln(1 + 2x) over 2x so the inner quantity matches the denominator. That factor tends to 1, leaving the coefficient 2.

\lim_{x \to 0} \frac{\ln(1 + 2x)}{x} = 2

Settled by matching the inner expression to the denominator.

Matching the inner expression

The standard limit is $\lim_{u \to 0}\frac{\ln(1+u)}{u} = 1$ , and it needs the same $u$ in both places. Here the inside carries a $2x$ , so manufacture a $2x$ underneath.

\frac{\ln(1+2x)}{x} = 2\cdot\frac{\ln(1+2x)}{2x}

\lim_{x \to 0}\frac{\ln(1+2x)}{x} = 2 \cdot 1 = 2

The general shape

The limit of ln(1 + ax) over x as x approaches 0 is a. It is the same coefficient the sine version produces, because near 0 both ln(1 + u) and sin u behave like u.

Why the standard limit is 1

It is a derivative in disguise. With $f(u) = \ln(1+u)$ and $f(0) = 0$ , the quotient is the alternate form of $f'(0)$ , and $f'(u) = \frac{1}{1+u}$ is $1$ at the origin.

L'Hopital confirms the whole thing in one pass, with the chain rule supplying the $2$ .

\lim_{x \to 0}\frac{\ln(1+2x)}{x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{\frac{2}{1+2x}}{1} = 2

The mistakes students make

Answering $1$ by pattern-matching without checking that the inner expression matches the denominator.
Splitting $\ln(1+2x)$ into $\ln 1 + \ln 2x$ . There is no log rule for a logarithm of a SUM.
Answering $\frac{1}{2}$ by inverting the coefficient. It multiplies rather than divides.

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 ln(1 + 2x) / x as x approaches 0?

It is $2$ .

What is the general rule?

$\lim_{x \to 0}\frac{\ln(1+ax)}{x} = a$ for any constant $a$ .

Can I split ln(1 + 2x)?

No. Logarithm rules apply to products and quotients, never to a sum inside the logarithm.