What is the limit of \( x^{1/x} \) as \( x \to \infty \)?

[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.

 

[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)}

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.

[)

O.K how do you prove the above ??

Or do you have a "limit law" that it is not true??

 

[Hurkyl]

*ahem* I refer you to your favorite introductory calculus textbook or similar reference material for the definition of continuity.

And should you be interested in the particulars of the construction and basic properties of a real exponential, look for a calculus/analysis text that covers that topic. (typically such a book would start with the construction of the natural logarithm via integration)

 

[evagelos]

*ahem* I refer you to your favorite introductory calculus textbook or similar reference material for the definition of continuity.

And should you be interested in the particulars of the construction and basic properties of a real exponential, look for a calculus/analysis text that covers that topic. (typically such a book would start with the construction of the natural logarithm via integration)

I did not ask for references ,neither did ask for a proof of the said law ,but i did ask for a particular proof using the ε-δ definition of a limit . A proof which does not exist in any books.

Now if anybody happens to know that proof i will be very glad to see it .

Of course if he/she wishes to do so

 

[HallsofIvy]

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.

Seriously, dude, learn some mathematics. NO function goes to "\infty/\infty. That may be what you get when you plug the target value into the fraction but that does not tell you what the limit is!

Would you say that \frac{x}{x} "goes to" \frac{\infty}{\infty}? It doesn't- for all x, \frac{x}{x}= 1 so its limit is 1.

Yes, ln(x) is unbounded. But it goes to infinity slowly, much slower than x. Sufficiently slower so that \lim_{x\to\infty}\frac{ln(x)}{x}= 0.

 

[ekcg]

laaa

 

[ekcg]

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.

\frac{ln(x)}{x} does in fact go to \frac{\infty}{\infty} when x tend to inf.
this is an indeterminate form. u can use l'hopital's rule to solve this easily.

easier way of doing this can be explained usingthe following example
lim x -->inf (x^5/x^4). lim x--> inf (x^5) is inf and lim x-->inf (x^4) is inf. it is inf because x^5 grows faster than x^4 when x tends to infinity.

so the higher the power of x, the faster it grows. one thing we have to understand is that e^x grows faster than ANY function of x and ln(x) grows slower than ANY function of x when x tends to infinity. so \frac{ln(x)}{x} when x tends to inf is just zero.

so both answers of indeterminate and zero is correct. inf/inf is just a way of expressing two functions are going towards infinity. it is an indeterminate value because we do not know how fast each functions are going towards infinity. and inf/inf is NOT always 1. in the example given by HallsofIvy. "limx->inf (x/x) is not inf/inf but 1." fact is it CAN BE expressed as inf/inf and l'hopital's rule can be used, which would give the answer 1. and this example gives an answer of 1 only because we know the two functions grows at the same rate (simply because they are the same function). so if the example was 2x/x. this would also be inf/inf when x-->inf. but the answer would not be 1 because the two functions are NOT growing at the same rate. and also simple questions like lim x-->inf (2x/x) would not require l'hopital's rule because we can instantly see that 2x grows twice as fast as x, or can easily simplify this with algebra. BUT using l'hopital's rule for this is perfectly fine.

i hope this clears up the argument between the answers of zero and inf/inf. they are both correct.

 

[Anonymous217]

3 year old thread necro-bumped for the second time.

 

[alphachapmtl]

Steiner's Problem
For what value of x is f(x)= x^(1/x) a maximum? The maximum occurs at x=e.
http://mathworld.wolfram.com/SteinersProblem.html

 

[Prove It]

The standard method of solving indeterminate forms is to apply a transformation to get \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty}.

In this case \displaystyle \lim_{x \to \infty}x^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left(x^{\frac{1}{x}}\right)}}

\displaystyle = \lim_{x \to \infty}e^{\frac{1}{x}\ln{x}}

\displaystyle = \lim_{x \to \infty}e^{\frac{\ln{x}}{x}}

\displaystyle = e^{\lim_{x \to \infty}\frac{\ln{x}}{x}} from the continutiy of the exponential function, this also gives the required form \displaystyle \frac{\infty}{\infty} so that you can use L'Hospital's Rule

\displaystyle = e^{\lim_{x \to \infty}\frac{\frac{1}{x}}{1}} by L'Hospital's Rule

\displaystyle = e^{\lim_{x \to \infty}\frac{1}{x}}

\displaystyle = e^0

\displaystyle = 1.

 

[HallsofIvy]

im very aware of what we're dealing with

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

You are using the terminology incorrectly. Nothing "goes to undeterminate". Just putting "infinity" into the formula gives an expression that is undeterminate but that says nothing about the limit. The limit itself, the value the function "goes to" is either a number or does not exist. There is no "in between".

 

[brydustin]

Proof:
http://www.wolframalpha.com/input/?i=x^%281%2Fx%29
QED