Find the limit as x approaches 0+ of (lnx)/x using L'hospitals rule.

Question

When I do this, I keep getting stuck at 1/0 when you plug back into the equation after doing l'hospital once.

Answer 1

Use L'hopitals rule to find the limit.

If you take the derivative of the top, and the derivative of the bottom

derivative of lnx = (1/x)

and the derivative of x = 1

You should be able to eliminate your problem with the 0.

Answer 2

So, why is ∞ not a valid answer?

Sometimes that is the limit.

lHospital does not guarantee that the limit is defined. It just allows one to make sure that an indeterminate form is avoided. 1/0 is not indeterminate -- it is infinite.

The only thing that bothers me is that

lnx/x -> -∞/0 = -∞
but 1/0 = +∞
when x = δ, a very small positive value.

any ideas on that?

Answer 3

Whoops, sorry about that. It does seem like you still wind up with 1/0.

Hmm... let's see when you graph it it looks like it's approaching -infinity so that's probably why you keep winding up with 1/0 since that will give you infinity technically.

Also, I don't believe there's any way to further simplify this problem so that you don't wind up with 1/0.

If you graph it you'll see what I'm talking about.

Answer 4

I am not sure L'Hopital's rule is relevant here because the limit of the top derivstive is -oo and of the bottom is 1 but if you try it with numbers and your calculator you will indeed get -oo
for example
ln .0001 = -9.21
-9.21/.0001 = -92103