Question

Why is 1^infinity an indeterminate form? Seems like 1 to any power would be 1.

Answer 1

this is a statement about limits, not about numbers

Answer 2

for example \((1+\frac{1}{n})^n\) is of the "form" \(1^{\infty}\) but it is one definition of \(e\)

Answer 3

That's exactly what I have had trouble with. In solving this limit the logic seems circular.

Answer 4

So we put it into an exponential form and do L'Hopital's rule where it becomes e^1. I'm just not satisfied with that being done in this way. It's saying, "this is indeterminate because it becomes e" and then "you see, it's e!" or maybe it's been too long since I've done limits...

Answer 5

you do not interpret a limit by its form. that is why the forms are called "indeterminate"
for example, you may want to argue that \(\frac{0}{0}=1\) because it is the same number top and bottom, or that \(\frac{0}{0}=0\) because there is a zero in the numerator, but \(\frac{0}{0}\) just means both the numerator and denominator approach 0, not that they are actually 0

Answer 6

Fair enough, makes sense, I'm looking at it from the backwards way basically.

Answer 7

as for \(\lim_{n\to \infty}(1+\frac{1}{n})^n\) it is some number that is not 1 as you can check for yourself.
take larger and larger values of \(n\) and see what you get. the fact that in the limit \((1+\frac{1}{n})=1\) does not mean you are raising one to the power of infinity
what you need is the precise definition of what a limit is, and then it will be more clear

Answer 8

No, it's just been a year since I took calculus and limits didn't really interest me because they seemed pretty simple, but I've started getting interested in understanding e^(ix) and why the integral of 1/x is lnx and things like that because they seem to just be thrown under the rug in school.

Answer 9

Err... So yeah, thanks for refreshing my memory! =D