# How do you find the limit of #(ln x)^2 / x # as x approaches infinity?

I would say that tends to zero.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of (ln x)^2 / x as x approaches infinity, we can use L'Hôpital's rule. By differentiating the numerator and denominator separately, we get (2ln x) / 1. As x approaches infinity, ln x also approaches infinity. Therefore, the limit is infinity.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the limit of # ( sqrt(x^2+11) - 6 ) / (x^2 - 25)# as x approaches 5?
- What is the limit as x approaches infinity of #6cos(x)#?
- What is the limit of #sqrt(x^3-2x^2+1)/[x-1]# as x goes to infinity?
- How do you find the x values at which #f(x)=(x-3)/(x^2-9)# is not continuous, which of the discontinuities are removable?
- How do you use the Squeeze Theorem to find #lim xcos(50pi/x)# as x approaches zero?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7