lim x^{1/x}, x-> infty

morphism

Right. OK.

JohnDuck

It makes perfect sense to me. What exactly is wrong with it?

ice109

at second glance the proof is fine but he still had to use an auxiliary theorem to prove the limit is 0

Pseudo Statistic

Seconded-- I mean, look at log(x); its derivative is 1/x where as x's derivative is always 1.
As you can see, log x's derivative is always less than x's derivative-- that is to say that x goes to infinity faster than log(x) would. (Equivalent to what L'Hopital would tell you I guess)
That's one way to look at it-- but morphism would've won me over with that elegant application of the squeeze theorem.

Take_it_Easy

REALLY?
:D

Maybe we have some misconception due to a different meaning of the word "indeterminate form".
If you mean that [tex] \frac{\infty}{\infty} [/tex] is always an indeterminate form OK!


It is an indeterminate form OK, but there are plenty of ways we can solve the indeterminateness.

Also [tex] \frac{x^2}{x} [/tex] and [tex] \frac{x}{x^2} [/tex] and [tex] \frac{x}{x} [/tex] are the same type of indeterminate limits as they are
[tex] \frac{\infty}{\infty} [/tex] but it all we know that they HAVE limit that are (respectively) [tex] \infty, 0, 1 [/tex] .


[tex] \frac{ln(x)}{x} [/tex] tends to 0.
I hate the De L'hopital tool but you can apply this and solve the matter yourself and see that it indeed goes to 0.

Hope this helps.

Take_it_Easy

Another proof.
It uses the elementry fact that [tex]e^y > y[/tex].
When y>0 you can apply the logaritm to that disequation and obtain
[tex]y > \ln(y)[/tex]
This hold for all [tex]y > 0[/tex].
Now when [tex]x \mapsto \infty[/tex] notice that
[tex]0 < \frac{\ln(x)}{x} = \frac{2\ln(x)}{2x}
=\frac{2\ln(\sqrt x)}{x} < \frac{2\sqrt x}{x} = \frac{2}{\sqrt x } \mapsto 0[/tex]

ice109

lol why did you bump this thread

Take_it_Easy

Yeah! I just misunderstood everything.
Now I read carefully there was no need to insist.
The reason is I thought you were not sure that limit went to 0 so I used my argument, but in the end I understand you are perfecly aware and I just misunderstood the matter of the topic.
I think I am not the only one who misunderstood.

My apologize and see you next time!

Ga San Wu

Hey guys I realise that this post is years late but I think its still useful for others to read ideas. I was looking at the start of the question which has

x^(1/x)

I was wondering whether there was an easier way to prove that this tends to 1 as x tends to infinity. Can we not just focus on this part first:

1/x

We can definitely show that this tends to 0 as x tends to infinity and so due to this no matter what

x^(1/x)

Will tend to 1 as its power tends to 0

Cyosis

Your argument is false. Using your logic what would the limit be of (e^x)^(1/x) when x tends to infinity?

Ga San Wu

I would have thought that this would tend to e
Maybe I should have explained myself a bit better as my explanation was kinda just for the first example.

Anything to the power of 1/x will tend to 1 as x tends to infinity
so as in your example the whole answer does not tend to 1
but the part with x^(1/x) does and so the answer is e

In a sense I am just looking at certain parts of the question before adding on any other part.

Sorry if my explanation is confusing! Obviously if I'm wrong please tell me too.

Cyosis

That explanation is incorrect as well, even though you find the correct answer. There is no x^(1/x) term in (e^x)^1/x. Remember (a^b)^c=a^(bc) and not a^(b^c).

Hurkyl

So you agree that e^x, to the power of 1/x, tends to e?

So you claim that e^x, to the power of 1/x, tends to 1?


We have a limit law that says

[tex]\lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}[/tex]

whenever both limits on the r.h.s. exist, and the limits are in the domain of the exponentiation function, and exponentiation is continuous in a neighborhood of that point.

[itex](+\infty)^0[/itex] isn't even in the domain of exponentiation! (If you don't know what an "extended real number is", then what I just said means that that's an indeterminate form)

Ga San Wu

D'oh! Just wrote out loads of stuff only to realise I clicked on report instead of new post!
Again I realise that I haven't explained myself properly which is causing misunderstandings, completely my fault.

To Cyosis:
Thanks for pointing out to me that I am wrong in that part. I had completely forgotten that rule for some reason, silly me! Although in your question (e^x)^(1/x) this just equals e^(x/x),I was wondering however whether you could show me an example where a variable with only ^1/x does not tend to 1. I'm guessing it would be something with different variables used like n^(1/x) where n grows a lot faster than x?

To Hurkyl:
Sorry I wasn't trying to claim that (e^x)^(1/x) tends to 1, for some reason I thought what Cyosis pointed out to me that (a^b)^c=a^(bc) and not a^(b^c).

P.S I don't want you guys to get the wrong impression of me, I'm not just trying to cause argument but get the grasp of all this so I can have a better understanding. Plus I think it's helping me to structure points better.

Whatever123

To the people arguing the limit of ln(x)/x as x ---> infinity... that's just 0. Using l'hoptial's rule yields 1/x/1 = 1/x which goes to 0 as x goes to infinity...

snipez90

Yeah dude you are about 3 years late.

evagelos

How, by using the ε-δ definition of a limit, can we prove the above??