Which statement correctly justifies lim_{x→0} ln(1+x) = 0?

Which statement correctly justifies lim_{x→0} ln(1+x) = 0?

• A. Because ln(1+x) tends to 0 as x→0 and ln is continuous at x=1 (ln(1)=0).
• B. Because ln(1+x) diverges as x→0.
• C. Because ln(1+x) tends to 1.
• D. Because ln has a derivative of 1 at x=0.

Answer: (A)

The limit is about what happens to the outside function when its input approaches a certain value. Here, as x approaches 0, the inside 1+x approaches 1. The natural logarithm is continuous at 1, so you can pass the limit through: lim_{x→0} ln(1+x) = ln(lim_{x→0}(1+x)) = ln(1) = 0. This is why the statement that ties the limit to ln(1) being 0 is the correct justification. The other ideas don’t fit: the expression does not diverge as x→0, the value it would approach is not 1 (it’s 0, since ln(1)=0), and using a derivative at 0 isn’t the direct reason the limit is 0 (the result follows from continuity, not the derivative).

The limit is about what happens to the outside function when its input approaches a certain value. Here, as x approaches 0, the inside 1+x approaches 1. The natural logarithm is continuous at 1, so you can pass the limit through: lim_{x→0} ln(1+x) = ln(lim_{x→0}(1+x)) = ln(1) = 0. This is why the statement that ties the limit to ln(1) being 0 is the correct justification.

The other ideas don’t fit: the expression does not diverge as x→0, the value it would approach is not 1 (it’s 0, since ln(1)=0), and using a derivative at 0 isn’t the direct reason the limit is 0 (the result follows from continuity, not the derivative).