Evaluate lim x→∞ (ln x)^2 / x.

Evaluate lim x→∞ (ln x)^2 / x.

• A. ∞
• B. -∞
• C. 1
• D. 0

Answer: (D)

As x grows large, the logarithm grows without bound but far more slowly than x itself. Even squaring ln x doesn’t change the fact that it is still overwhelmed by the linear growth of x, so the ratio (ln x)^2 / x tends to zero. A clean way to see this is to use L’Hôpital’s rule. The form is ∞/∞, so differentiate top and bottom: derivative of (ln x)^2 is 2(ln x)(1/x) = 2 ln x / x, and derivative of x is 1. Now you have lim x→∞ (2 ln x)/x, still ∞/∞, so apply L’Hôpital again: derivative of 2 ln x is 2/x, and derivative of x is 1, giving lim x→∞ 2/x = 0. Therefore, the limit is 0.

As x grows large, the logarithm grows without bound but far more slowly than x itself. Even squaring ln x doesn’t change the fact that it is still overwhelmed by the linear growth of x, so the ratio (ln x)^2 / x tends to zero.

A clean way to see this is to use L’Hôpital’s rule. The form is ∞/∞, so differentiate top and bottom: derivative of (ln x)^2 is 2(ln x)(1/x) = 2 ln x / x, and derivative of x is 1. Now you have lim x→∞ (2 ln x)/x, still ∞/∞, so apply L’Hôpital again: derivative of 2 ln x is 2/x, and derivative of x is 1, giving lim x→∞ 2/x = 0. Therefore, the limit is 0.