What is the value of $\sqrt[\infty]{\infty}$?

[Jonathan G]

Homework Statement

$$\sqrt[\infty]{\infty}$$

Homework Equations

n/a

The Attempt at a Solution

I remember the answer being either 1 or 0 but I can't remember right now and seeking guidance please. I'm pretty sure it is 1 but I wan't to be 100% sure. I'm working with sequences and series right now.

 

[Bohrok]

You want to find this limit
$$\lim_{x\rightarrow\infty} x^{\frac{1}{x}}$$

It's of indeterminate form ∞⁰. Can you rewrite it so you can use L'Hôpital's rule?

 

[Jonathan G]

OK, yeah that helped...The answer is 0. Thanks

 

[njama]

You mean:

$$\lim_{n\rightarrow +\infty}\sqrt[n]{n}=\lim_{n\rightarrow +\infty}n^{1/n}=1$$

As n goes to +infinity 1/n -> 0 and n⁰=1

You can convince your self by putting large values (1000, 10000, etc...) for n in the calculator.

 

[Jonathan G]

hmm...so i was initially correct. =P dang..so i did something wrong when doing L'Hopital's rule. =/

 

[Mark44]

OK, yeah that helped...The answer is 0. Thanks

Maybe, or maybe not. An indeterminate form is called that because you can't determine beforehand just what the value will be.

 

[Jonathan G]

Well, I tried using L'Hopital's rule and I ended up getting 0 (zero) but then I tried njama's way and I got 1. And I remember back in HS my calc teacher mentioning something about this problem but I just can't seem to remember the details. I'll continue trying to determing this problem. I'm sure It'll come to me. ^_^

 

[Bohrok]

You should get 0 for the part you use with L'hôpital's rule if you rewrite the problem correctly, which you probably did. The limit 0 should be the exponent on e, e⁰ = 1. Show what you've tried so far.

 

[Jonathan G]

ahh yes...thank you Bohrok..I did get 0 but I forgot that I had initially used a log. Thanks for clearing it up. ^_^

 

[Elucidus]

Quick comment:

Many of my students assume that indeterminate powers of the form $\infty^0$ always have a limit of 1. This is not true in general.

The example $\lim_{x \rightarrow \infty}x^{1/x}$ does have a limit of 1, but

$$\lim_{x \rightarrow \infty}(\sinh 4x)^{1/x}}= e^4$$.

--Elucidus

 

[Jonathan G]

Yes, I am well aware of this. I know when you get infinity to the power of zero the solution is not automatically 1 just because of the power, hence the reason it is an indeterminate form. The same goes with infinity divided by infinity or zero divided by zero. Right away one thinks..oh its impossible when in fact with just a little bit of simple mathematics it can easily come to a solution.

I know you probably were not stating this directly at me and simply letting everybody see but meh.

 

[Elucidus]

Yes, I am well aware of this. I know when you get infinity to the power of zero the solution is not automatically 1 just because of the power, hence the reason it is an indeterminate form. The same goes with infinity divided by infinity or zero divided by zero. Right away one thinks..oh its impossible when in fact with just a little bit of simple mathematics it can easily come to a solution.

I know you probably were not stating this directly at me and simply letting everybody see but meh.

My comment was directed at whoever may read this thread and misconstrue that $$\sqrt[\infty]{\infty}$$ as being 0 or 1 and not reading it thoroughly. L'Hospital's Rule gets so abused it's frequently disheartening. No criticism intended

--Elucidus

 

[njama]

My comment was directed at whoever may read this thread and misconstrue that $$\sqrt[\infty]{\infty}$$ as being 0 or 1 and not reading it thoroughly. L'Hospital's Rule gets so abused it's frequently disheartening. No criticism intended

--Elucidus

I don't know why you think that it is not 1 when it goes to positive infinity?

 

[HallsofIvy]

I don't know why you think that it is not 1 when it goes to positive infinity?

njama, I have no idea what this is supposed to mean. You used the pronoun "it" twice with no clear antecedent- and, I think, to mean two different things.

And, Jonathon, you started by asking "what is $\sqrt[\infty]{\infty}$" and then said that you are 'well aware" that the value of such an "indeterminate" form will depend on what precise function you are taking the limit of. Do you mean you were "well aware" that your original question was nonsense?

 

[njama]

njama, I have no idea what this is supposed to mean. You used the pronoun "it" twice with no clear antecedent- and, I think, to mean two different things.

Sorry, mate. "It" was referring to the limit. To be clearer, why the limit can't be 1?

 

[Jonathan G]

To be more precise for HallsofIvy the actual problem was as follows:

Determine whether the sequence converges or diverges. If it converges, find the limit.

the problem was $$a_{n}$$ = ($$2^{1+3n}$$) $$^{1/n}$$

 

[Mark44]

To be more precise for HallsofIvy the actual problem was as follows:

Determine whether the sequence converges or diverges. If it converges, find the limit.

the problem was $$a_{n}$$ = ($$2^{1+3n}$$) $$^{1/n}$$

That's what you should have put in your first post, right where it says Homework Statement

 

[njama]

$$a_n=2^{(1+3n)/n}=2^{1/n+3}=2^{1/n}*2^{3}$$

 

[Jonathan G]

njama

$$a_{n}$$ = ( $$2^{1+3n}$$ ) $$^{1/n}$$ ,the power of 1/n is the nth-root of the 2^(1+3n) not the way u stated it.

sorry if I messed you up.

 

[Elucidus]

njama is correct in that

$$(2^{1+3n})^{1/n} = 2^{3+1/n}$$

The limit as n approaches infinity can be obtained either by using logarithms and L'Hospital's rule or by exploiting the continuity of an exponential.

Either way the answer is 8.

This is an example of an indeterminate power of the form [itex]\infty^0[/tex], which was originally described as of the type $$\sqrt[\infty]{\infty}$$ (which is a bit of a misnomer since it isn't necessarily always an integral root).<br /> <br /> --Elucidus[/itex]