Question

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Answer 1

\[\lim_{n \rightarrow \infty} \ln (1/n) = -\infty\]
Steps please.

Answer 2

I formulated my question in a confusing way last time.

Answer 3

ln(1/n) = ln(1) - ln(n) = 0 - ln(n) = -ln(n)
so lim(as n->inf) of -ln(n) = -ln(inf) = -inf

Answer 4

Thank you! I used hoptal's rule and got \[1/n(-1/n ^{2}) = -1/n\]
This gave me 0 and is obviously wrong. Can you tell me what I did wrong so I don't repeat the same mistake?

Answer 5

L'hopital's rule is used when you have an indeterminate case... inf/inf, inf*inf, 0/0, etc... we don't have that in this case...

Answer 6

But ln 0 is an error but that doesn't fall in the case, right?

Answer 7

right, it doesn't because ln 0 is not defined, not indeterminate.

Answer 8

Okay, thank you very much!

Answer 9

yw, np