# L'hopital's rule/ possible indeterminate form question

## Homework Statement

limit as n approaches infinity of (ln(1/n) / n)

## The Attempt at a Solution

Would it be valid to exponentiate (not sure if you can make that into a verb or not) the top and bottom, yielding

(1/n) / e^n

Then, if I take the limit I get 0/infinity. Is it valid to say the limit is zero? Is 0 over infinity indeterminate?

## Answers and Replies

Dick
Homework Helper
You can 'exponentiate' but you sure can't do it like that. e^(a/b) IS NOT e^a/e^b. Why don't you just try using l'Hopital without the bogus exponentiation?

I'm confused by what is an indeterminate form and what isn't. Before using l'Hopital's rule, if you take the limit of the function as is, you have a limit that does not exist over infinity. I wasn't sure if I could use it at that point or not.

For instance, I know that inf - inf, 0/0, inf/inf, inf^0, 1^inf, and 0^0 are all indeterminate. What if a function evaluated at some point doesn't result in zero or infinity, but simply does not exist at that point?

Dick
Homework Helper
lim ln(1/n) is -infinity, yes? Since 1/n->0. n->infinity. So the form of the limit is -infinity/infinity. That's pretty indeterminant. It's like -2n/n as n->infinity. You can certainly use l'Hopital on that.

That makes sense, thank you sir.

gb7nash
Homework Helper
another way to look at it is that ln(1/n) = -ln(n) since ln(ab) = blna.