Question

Can anyone offer a good explanation / reasoning for this?

(see below)

Answer 1

None at all.

Answer 2

\[\large \lim_{x \rightarrow ∞} (1)^x\] is indeterminate?

Answer 3

I'm pretty sure it's not, that limit equals 1. If it were -1 instead of 1 it'd be indeterminate.

Answer 4

it is indeterminate form.. any value which has power infinity is indeterminate.

Answer 5

"it is indeterminate form.. any value which has power infinity is indeterminate." @nubeer , that's what I'm getting, but I want to know why that is the case.

Answer 6

Depending on how you arrived to that 1 you might be able to find a value for the limit, check out some examples here: http://www.vitutor.com/calculus/limits/one_infinity.html

Answer 7

hmm you can say its something like preset or defination .. maybe i am not sure maybe someone else cna explain.

Answer 8

At this stage of your studies, the explanation is this is one of the special cases. Other special cases are \[\infty ^{0} and 0^{0}\]

Answer 9

\[\lim_{x \rightarrow \infty}a^b\] where a->1 and b-> inf
is an indetermination. It can usually be solved, but you'll learn about that later.
\[\lim_{x \rightarrow \infty} 1^{\infty}\]if given directly is simply 1.

Answer 10

Re: "At this stage of your studies, . . ."

I am a calculus teacher, and I'm looking for a more convincing reason for my students.

Answer 11

Also \[\frac{ 0 }{ 0 }\]

Answer 12

Zero times infinity,
and infinity - infinity
I'm sorry, that is what my teacher told me in Cal 1, that these are special cases.

Answer 13

Thing is, there's no reason it can't be done; as you know it can definitely be done, just at a higher level. You could just tell them they'll learn to do these later.

Answer 14

I can explain the reasons behind the other special cases, 0/0. ∞/∞, ∞^0, 0^0.
But when students are so used to hearing that "1 raised to any power is still 1." it's hard to come up with a good reason why 1 to a ridiculously unlimitedly large number would be anything other than still 1.

Just saying, well it's to the infinite power (which it isn't because it's a limit), and that makes it special sounds lame and a cop-out. My students expect much better from me.

Answer 15

Intuitively, it seems indeterminate because while the power goes to infinity, the whole thing goes to 1

Answer 16

1 to any power, no matter how astronomical, is 1. However, when you take a limit you don't have 1, you have a number that's close to 1, but not quite. If you raise 0.999... or 1.001... to a very large number, a number that approaches infinity faster than the other number approaches 1, you can definitely get a different result.

Answer 17

Of course, "approaches infinity faster than the other number approaches 1" has no rigor whatsoever, but it ought to do the trick.

Answer 18

Yeah, I can understand that in a^x, a approaches 1 as x becomes infinite, but if the 1 is definitely 1, then why not \(1^∞=1?\)

I'm beginning to think that it's simply not true.

From here: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityII.aspx
it is shown e^x going infinite as x becomes infinite, so why not 1^x?

I'm going to try finding it in that book again. Maybe it's a typo.

Answer 19

Ok, I found the reference. It is in ARCO AP Mathematics, 4th ed. p.90 "The indeterminate forms 0^0, ∞^0 and 1^∞ can be handled by first taking logarithms . . ."

That's fine, but it still doesn't explain why 1^∞ is indeterminate in the first place.

Answer 20

No, it's certainly true. The value of a^x (a is a constant, not a limit) when x goes to infinity is
-infinity for a>1
-1 for a=1
-zero for -1<a<1
-non existant for a=-1
-minus infinity for a<-1
Now suppose that instead of a constant we have a limit for a. In the discontinuities (1 and -1), we can expect trouble. We know that a approaches 1, but we don't know if a approaches 1 from above (where the limit should be infinity) or from below (then the limit should be 0). a could even move above and below 1. For that reason, that limit could take almost any value, and we need to get some more information ( http://www.vitutor.com/calculus/limits/one_infinity.html) before we solve it.

Answer 21

Working link: http://www.vitutor.com/calculus/limits/one_infinity.html

Answer 22

Ah, I think I see what's going on. It's not saying that 1^∞ is itself indeterminate, but getting 1^∞ as a limit of a function as the variable approaches some other value is not a determined limit.

Answer 23

Perhaps I misread what the book was saying. It's a brief study-guide style book and just has lists of facts with little explanation.

\[\large \lim_{x \rightarrow a}f(x)=1^∞ \]
is not the same as
\[\large\lim_{x \rightarrow ∞}1^x=1\]

Thank you for the link, @jirafa , that clarified the difference in meaning.

Answer 24

You're welcome

Answer 25

On the phone with my buddy a Phd. I'm understanding, it is said to indeterminate in early classes, but later you find answer by using L'hopital

Answer 26

Yes, in the book where I read that, it is in the section right after L'Hopital's rule.

It only applies if the limit of something else appears to be 1^∞, but 1^∞ itself is equal to 1.