As x approaches 0, what is the limit of ln(1+x)/x?

As x approaches 0, what is the limit of ln(1+x)/x?

• A. 0
• B. 1/2
• C. 2
• D. 1

Answer: (D)

For small x, ln(1+x) behaves like x. The ratio ln(1+x)/x then compares two quantities that are nearly the same, so it should approach 1. A precise way to see this is to use the series expansion: ln(1+x) = x - x^2/2 + x^3/3 - ..., so dividing by x gives 1 - x/2 + x^2/3 - ..., which tends to 1 as x → 0. Another route is L'Hôpital's rule: the limit of ln(1+x)/x as x → 0 is the limit of its derivatives, which is (1/(1+x))/1 = 1/(1+x) → 1.

For small x, ln(1+x) behaves like x. The ratio ln(1+x)/x then compares two quantities that are nearly the same, so it should approach 1.

A precise way to see this is to use the series expansion: ln(1+x) = x - x^2/2 + x^3/3 - ..., so dividing by x gives 1 - x/2 + x^2/3 - ..., which tends to 1 as x → 0.

Another route is L'Hôpital's rule: the limit of ln(1+x)/x as x → 0 is the limit of its derivatives, which is (1/(1+x))/1 = 1/(1+x) → 1.