# Lim x->0+ and l'Hospital Rule

1. Jul 10, 2010

### phillyolly

1. The problem statement, all variables and given/known data

lim(x→0^+)(lnx/x)

3. The attempt at a solution
First, what does x->0+ mean? Is it positive infinity?
Why lim(x→0^+) (lnx)=-∞?

= -∞ * (1/0+)
= -∞* ∞
= -∞?
Why cannot I use l'Hospital Rule?

2. Jul 10, 2010

### Staff: Mentor

It means that x approaches zero from the right (positive values close to zero).
Look at the graph of y = ln(x). The domain is {x | x > 0}. As x gets closer to zero, y gets more and more negative.
What you have above is mostly incorrect. 1/0 is not a number, so you can't use it in calculations. $$\lim_{x \to 0^+} \frac{ln(x)}{x} = \lim_{x \to 0^+} \frac{1}{x}ln(x)$$
The first factor gets larger and larger without bound; the second factor gets more and more negative without bound. As a result the product's limit is -∞.
Read the fine print in L'Hopital's Rule. There are certain conditions that must be satisfied before you can use it.

3. Jul 10, 2010

### phillyolly

That's a great explanation. Thank you a lot.