L'hopital's rule/ possible indeterminate form question

[Sentience]

Homework Statement

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

Homework Equations

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?

 

[Dick]

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?

 

[Sentience]

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.

 

[Sentience]

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]

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.

 

[Sentience]

That makes sense, thank you sir.

 

[gb7nash]

another way to look at it is that ln(1/n) = -ln(n) since ln(a^b) = blna.