Why Does l'Hospital's Rule Fail for lim(x→0^+) (lnx/x)?

[phillyolly]

Homework Statement

lim(x→0^+)(lnx/x)

The Attempt at a Solution

First, what does x->0+ mean? Is it positive infinity?
Why lim(x→0^+) (lnx)=-∞?
Why the answer is

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

 

[Mark44]

Homework Statement

lim(x→0^+)(lnx/x)

The Attempt at a Solution

First, what does x->0+ mean? Is it positive infinity?

It means that x approaches zero from the right (positive values close to zero).

Why lim(x→0^+) (lnx)=-∞?

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.

Why the answer is

= -∞ * (1/0+)
= -∞* ∞
= -∞?

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 -∞.

Why cannot I use l'Hospital Rule?

Read the fine print in L'Hopital's Rule. There are certain conditions that must be satisfied before you can use it.

 

[phillyolly]

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