It's been a while, but let's say:

y = lim x --> 0 (x+1)^(1/x)

then ln y = lim ln[(x+1)^(1/x)] = lim (1/x)*ln(x+1)

as x --> 0 this gives the indeterminate 0/0, so use L'Hôpital's rule:

ln y = lim (1/(1+x))/1

as x --> 0 the limit is 1, so ln y = 1; ergo, y = e.

I think? Now my question is how does lim x-->inf 1^x equal 0? Did you mean 1/x?

It's been a while, but let's say:y = lim x --> 0 (x+1)^(1/x)then ln y = lim ln[(x+1)^(1/x)] = lim (1/x)*ln(x+1)as x --> 0 this gives the indeterminate 0/0, so use L'Hôpital's rule:ln y = lim (1/(1+x))/1as x --> 0 the limit is 1, so ln y = 1; ergo, y = e.I think? Now my question is how does lim x-->inf 1^x equal 0? Did you mean 1/x?

Yeah, I meant i^(1/x).

But you misunderstood my question, I know how to prove both limits, but why do they come out to different things? Why does 1^inf not equal 1?

Oh, oops. As to why 1^inf is indeterminate and not 1, well, it relies on knowing another indeterminate form.

if y = 1^inf, then ln y = ln 1^inf and we can rewrite y = e^(ln 1^inf) = e^(inf*ln 1) = e^(inf*0).

Since inf*0 is indeterminate, then y, which was 1^inf, is indeterminate as well.

As for why inf*0 is defined as such and not 0, well, this is a bad time for me to have a Prof. Layton avatar.

Aha, that makes sense. Thank you very much!