Evaluate the limit as x approaches 0 of ln(1 + x) divided by x.

Evaluate the limit as x approaches 0 of ln(1 + x) divided by x.

• A. 0
• B. 1
• C. -1
• D. 1/2

Answer: (B)

The limit tests how ln(1+x) behaves relative to x right near zero. Since ln(1+x) is differentiable at 0 and its derivative there is 1/(1+0) = 1, the standard limit statement says that the rate of change of ln(1+x) at 0 equals the ratio [ln(1+x) − ln(1)]/(x − 0). Because ln(1) = 0, this is just ln(1+x)/x, and it must approach 1 as x approaches 0. Another way to see it is the series ln(1+x) = x − x^2/2 + ..., which when divided by x gives 1 − x/2 + ..., tending to 1. L'Hôpital’s rule yields the same result: the derivative of the numerator is 1/(1+x) and the derivative of the denominator is 1, which at x → 0 gives 1. So the limit is 1.

The limit tests how ln(1+x) behaves relative to x right near zero. Since ln(1+x) is differentiable at 0 and its derivative there is 1/(1+0) = 1, the standard limit statement says that the rate of change of ln(1+x) at 0 equals the ratio [ln(1+x) − ln(1)]/(x − 0). Because ln(1) = 0, this is just ln(1+x)/x, and it must approach 1 as x approaches 0. Another way to see it is the series ln(1+x) = x − x^2/2 + ..., which when divided by x gives 1 − x/2 + ..., tending to 1. L'Hôpital’s rule yields the same result: the derivative of the numerator is 1/(1+x) and the derivative of the denominator is 1, which at x → 0 gives 1. So the limit is 1.