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.
Settled by substitution to a standard logarithm limit.
Match the denominator
The standard limit is . Here , so the denominator wants to be .
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 at , which is evaluated at 0, namely 1. Equivalently, the Maclaurin series is , whose leading term is .
The series view also explains the size estimate: for small , 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?
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 runs to as , so there is no indeterminate form and no finite limit. The 1 inside is what keeps the logarithm near zero.