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

[dtkyi]

This seems to be an obvious problem, but for some reason I'm stumped. So what is the limit of x^{1/x} as x approaches infinity? I know that limit of (1+1/x)^{x} = e (as x-> infty), and limit of (1+x)^{1/x} = e (as x-> 0), but this is slightly different.

Any help is appreciated :)

 

[ice109]

its an indeterminate form, do l'hospitale

 

[mathman]

One way to solve it is by observing that;

x^1/x=e^lnx/x.

Since lnx/x -> 0 as x ->oo, the answer you want is 1.

 

[ice109]

One way to solve it is by observing that;

x^1/x=e^lnx/x.

Since lnx/x -> 0 as x ->oo, the answer you want is 1.

no lim lnx/x -> oo/oo as x->oo , you still get an indeterminate form. but i realize applying l'hospitale directly to the first expression is pointless

 

[Office_Shredder]

well, no ice, lim lnx/x as x goes to infinity is just 0. The fact that you can't split it up doesn't change that fact

 

[ice109]

well, no ice, lim lnx/x as x goes to infinity is just 0. The fact that you can't split it up doesn't change that fact

what? how is that even true? since when is the ln function bounded?

 

[morphism]

what? how is that even true? since when is the ln function bounded?

We're dealing with ln(x)/x as a single entity, not just ln(x)...

 

[dtkyi]

Thanks guys for clearing this up. It seems the main idea is to rewrite x^{1/x} into the form e^{1/x * ln x}, so that the exponent -> 0 as x -> infinity, and thus e^{0} = 1.

 

[ice109]

We're dealing with ln(x)/x as a single entity, not just ln(x)...

im very aware of what we're dealing with

$$\frac{ln(x)}{x}$$ does not go to zero at infinity, it goes to undeterminate.

 

[morphism]

im very aware of what we're dealing with

$$\frac{ln(x)}{x}$$ does not go to zero at infinity, it goes to undeterminate.

No. It goes to zero.

Here's a proof that doesn't use L'Hopital.

If x>0, then $\ln x < x$, so $(\ln{\sqrt{x}}) / \sqrt{x} < 1$.

So if x>1, then:
[tex]0 < \frac{\ln x}{x} = \frac{2\ln{(\sqrt{x})}}{\sqrt{x}\sqrt{x}} < \frac{2}{\sqrt{x}}[/itex]

A quick application of the squeeze/sandwich theorem now proves that ln(x)/x goes to zero.

And what exactly do you mean when you say "goes to undeterminate" anyway? The very fact that you use L'Hopital on lim ln(x)/x to get lim ln(x)/x = lim 1/x = 0 proves that the limit is 0.

 

[ice109]

No. It goes to zero.

Here's a proof that doesn't use L'Hopital.

If x>0, then ln(x) < x, so ln(sqrt(x))/sqrt(x) < 1.

So if x>1, then:
0 < ln(x)/x = 2ln(sqrt(x))/(sqrt(x) * sqrt(x)) < 2/sqrt(x)

A quick application of the squeeze/sandwich theorem now proves that ln(x)/x goes to zero.

And what exactly do you mean when you say "goes to undeterminate" anyway? The very fact that you use L'Hopital on lim ln(x)/x to get lim ln(x)/x = lim 1/x = 0 proves that the limit is 0.

i don't understand your proof, but what i mean is that as the expression stands $$\frac{ln(x)}{x}$$ goes the the undeterminate form $$\frac{\infty}{\infty}$$ and that l'hospitale has to be applied to see that the limit is zero. the initial point was that the rewriting of the expression using logs didn't save him any steps because still had to apply l'hospitale to figure out the limit. i then realized that applying l'hospitale to original expression was pointless.

 

[morphism]

i don't understand your proof, but what i mean is that as the expression stands $$\frac{ln(x)}{x}$$ goes the the undeterminate form $$\frac{\infty}{\infty}$$ and that l'hospitale has to be applied to see that the limit is zero. the initial point was that the rewriting of the expression using logs didn't save him any steps because still had to apply l'hospitale to figure out the limit. i then realized that applying l'hospitale to original expression was pointless.

The point is: you do not need L'Hopital to do that limit.

 

[ice109]

The point is: you do not need L'Hopital to do that limit.

you still used squeeze theorem? direct evaluation leads to an indeterminate form. who cares how you rectify the problem, obviously the limit of the expression as it stands is not just 0

seriously dude rewrite your proof in tex so i can understand it

 

[morphism]

who cares how you rectify the problem, obviously the limit of the expression as it stands is not just 0

How can you still say that?!

 

[ice109]

How can you still say that?!

omg dude seriously its not that difficult, what is the limit of $$\frac{1}{x}$$ as x approaches infinity? in the same sense as that limit is zero the other limit is not, you need an auxiliary scheme to prove it is zero, where this one is obvious by direct inspection. this is seriously down to semantics now. the end.

 

[morphism]

omg dude seriously its not that difficult, what is the limit of $$\frac{1}{x}$$ as x approaches infinity? in the same sense as that limit is zero the other limit is not, you need an auxiliary scheme to prove it is zero, where this one is obvious by direct inspection. this is seriously down to semantics now. the end.

You obviously don't know what the definition of a limit is.

$$\lim_{x \to \infty} \frac{\ln{x}}{x} = 0$$

It's absolutely NOT $\frac{\infty}{\infty}$.PS. I tex'd my proof.

 

[ice109]

yea terrible proof, and i'll just wait for someone else to refute you because I'm exasperated

 

[morphism]

Right. OK.

 

[JohnDuck]

yea terrible proof, and i'll just wait for someone else to refute you because I'm exasperated

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

 

[ice109]

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

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

 

[Pseudo Statistic]

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

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]

im very aware of what we're dealing with

$$\frac{ln(x)}{x}$$ does not go to zero at infinity, it goes to undeterminate.

REALLY?
:D

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


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

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


$$\frac{ln(x)}{x}$$ 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 $$e^y > y$$.
When y>0 you can apply the logaritm to that disequation and obtain
$$y > \ln(y)$$
This hold for all $$y > 0$$.
Now when $$x \mapsto \infty$$ notice that
$$0 < \frac{\ln(x)}{x} = \frac{2\ln(x)}{2x} =\frac{2\ln(\sqrt x)}{x} < \frac{2\sqrt x}{x} = \frac{2}{\sqrt x } \mapsto 0$$

 

[ice109]

lol why did you bump this thread

 

[Take_it_Easy]

what i mean is that as the expression stands $$\frac{ln(x)}{x}$$ goes the the undeterminate form $$\frac{\infty}{\infty}$$ and that l'hospitale has to be applied to see that the limit is zero. the initial point was that the rewriting of the expression using logs didn't save him any steps because still had to apply l'hospitale to figure out the limit. i then realized that applying l'hospitale to original expression was pointless.

lol why did you bump this thread

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 realize 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]

I would have thought that this would tend to e

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

Anything to the power of 1/x will tend to 1 as x tends to infinity

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


We have a limit law that says

$$\lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}$$

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.

$(+\infty)^0$ 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 realize I clicked on report instead of new post!
Again I realize 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]

We have a limit law that says

$$\lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}$$

)

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

 

[Hurkyl]

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

You can't, because it's not true.


Once you get the full statement of the theorem, please reference your favorite textbook or similar material for the definition of continuity and a statement about what domains upon which one might define an exponentiation operation, and where it might be continuous.