Compute the limit lim_{x->0} ln(1+x) / x.

Compute the limit lim_{x->0} ln(1+x) / x.

• A. 0
• B. 1
• C. ∞
• D. -1

Answer: (B)

Think about how ln(1+x) behaves when x is very small. It behaves almost like a straight line with slope 1 at the point x = 0, so the ratio ln(1+x)/x approaches that slope. A direct way to see this is by L’Hôpital’s rule: as x → 0, both numerator and denominator go to 0, so differentiate top and bottom to get lim x→0 (1/(1+x)) / 1 = 1/(1+0) = 1. Another view is the Taylor expansion: ln(1+x) = x − x^2/2 + …, so ln(1+x)/x = 1 − x/2 + … → 1 as x → 0. This limit exists from both sides (within the domain x > −1) and equals 1.

Think about how ln(1+x) behaves when x is very small. It behaves almost like a straight line with slope 1 at the point x = 0, so the ratio ln(1+x)/x approaches that slope.

A direct way to see this is by L’Hôpital’s rule: as x → 0, both numerator and denominator go to 0, so differentiate top and bottom to get lim x→0 (1/(1+x)) / 1 = 1/(1+0) = 1.

Another view is the Taylor expansion: ln(1+x) = x − x^2/2 + …, so ln(1+x)/x = 1 − x/2 + … → 1 as x → 0.

This limit exists from both sides (within the domain x > −1) and equals 1.